Power Series practice Problems with solutions

These power series problems cover every major topic you will encounter in Calculus II: finding the radius of convergence and interval of convergence using the Ratio Test and Root Test, representing functions as power series via geometric series manipulation, applying term-by-term differentiation and integration to power series, and constructing Taylor and Maclaurin series. Problems progress from straightforward convergence checks through partial-fraction decompositions, endpoint analysis, and series-based approximation of definite integrals.

Contents5 sections

Radius and Interval of Convergence

Determining where a power series converges is the foundational skill of the topic. The problems below build from single-step Ratio Test applications to cases requiring careful endpoint testing and the Root Test, reinforcing the three possible convergence behaviors: convergence at a single point, convergence on a finite interval, and convergence for all real numbers.

Problem 1: Ratio Test (Polynomial Coefficients)

Easy

Consider the power series centered at $x = 0$ :

\sum_{n=0}^{\infty} \frac{n+1}{3^n}\, x^n

Apply the Ratio Test to find the radius of convergence $R$ .
Test the endpoints $x = 3$ and $x = -3$ to determine the interval of convergence.

Hint

Form the ratio

\left|\dfrac{a_{n+1}}{a_n}\right|

and simplify. Remember that the Ratio Test gives

R = 1/L

where

L = \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|

. At the endpoints, decide whether the resulting series converges or diverges by comparing to a known test.

View Solution

Solution to part 1:

Let $a_n = \dfrac{n+1}{3^n} x^n$ . Compute:

\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(n+2)\,x^{n+1}}{3^{n+1}} \cdot \frac{3^n}{(n+1)\,x^n}\right| = \frac{n+2}{3(n+1)}\,|x| \to \frac{|x|}{3} \quad \text{as } n\to\infty.

The Ratio Test requires the limit to be less than 1, so $\dfrac{|x|}{3} < 1$ , giving $\boxed{R = 3}$ .

Solution to part 2:

At $x = 3$ : the series becomes $\displaystyle\sum_{n=0}^{\infty}(n+1)$ , which diverges (terms do not tend to zero).

At $x = -3$ : the series becomes $\displaystyle\sum_{n=0}^{\infty}(-1)^n(n+1)$ , which also diverges for the same reason.

Therefore the interval of convergence is $(-3,\, 3)$ .

Problem 2: Ratio Test (Non-Zero Center with Endpoint Analysis)

Medium

Determine the interval of convergence of the power series centered at $a = 2$ :

\sum_{n=1}^{\infty} \frac{(-1)^n}{n\cdot 5^n}\,(x-2)^n

Find the radius of convergence $R$ using the Ratio Test.
Check both endpoints and state the full interval of convergence, specifying whether each endpoint is included.

Hint

After finding

R = 5

, the interval is centered at

x = 2

, so the endpoints are

x = 2 \pm 5

. At one endpoint you get a

p

-series; at the other, an alternating series. Apply the appropriate convergence test to each.

View Solution

Solution to part 1:

\left|\frac{a_{n+1}}{a_n}\right| = \frac{1}{(n+1)\cdot 5^{n+1}}\cdot\frac{n\cdot 5^n}{1}\cdot|x-2| = \frac{n}{5(n+1)}|x-2| \to \frac{|x-2|}{5}.

Convergence requires $|x - 2| < 5$ , so $R = 5$ . The open interval is $(-3,\, 7)$ .

Solution to part 2:

At $x = 7$ (so $x - 2 = 5$ ):

\sum_{n=1}^{\infty} \frac{(-1)^n}{n\cdot 5^n}\cdot 5^n = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}.

This is the alternating harmonic series, which converges by the Alternating Series Test.

At $x = -3$ (so $x - 2 = -5$ ):

\sum_{n=1}^{\infty} \frac{(-1)^n}{n\cdot 5^n}\cdot(-5)^n = \sum_{n=1}^{\infty} \frac{(-1)^n(-1)^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n}.

This is the harmonic series, which diverges.

The interval of convergence is $(-3,\, 7]$ .

Problem 3: Root Test and Radius of Convergence

Hard

Consider the power series:

\sum_{n=1}^{\infty} \left(\frac{n}{n+1}\right)^{n^2} x^n

Apply the Root Test to determine the radius of convergence.
Explain why the Ratio Test is inconvenient here and what the Root Test reveals about the coefficient behavior.

Hint

For the Root Test, compute

\sqrt[n]{|a_n|} = \left(\dfrac{n}{n+1}\right)^n \cdot |x|

. Recall the fundamental limit

\lim_{n\to\infty}\left(1 + \dfrac{1}{n}\right)^n = e

, and write

\dfrac{n}{n+1} = \dfrac{1}{1 + 1/n}

View Solution

Solution to part 1:

Let $a_n = \left(\dfrac{n}{n+1}\right)^{n^2} x^n$ . Apply the Root Test:

\sqrt[n]{|a_n|} = \left(\frac{n}{n+1}\right)^n |x| = \frac{|x|}{\left(1 + \frac{1}{n}\right)^n} \to \frac{|x|}{e}.

Convergence requires $\dfrac{|x|}{e} < 1$ , giving $|x| < e$ . Therefore $\boxed{R = e}$ .

Solution to part 2:

The Ratio Test would require computing $\left(\dfrac{n}{n+1}\right)^{n^2} / \left(\dfrac{n+1}{n+2}\right)^{(n+1)^2}$ , an expression with unwieldy exponent differences. The Root Test handles the $n^2$ exponent cleanly by reducing it to an $n$ -th power, immediately invoking the classic limit for $e$ . This illustrates that the Root Test is superior when coefficients involve $n$ -th powers or super-exponential decay.

Representing Functions as Power Series

A key application of power series is rewriting rational and algebraic functions as infinite polynomial-like expressions using the geometric series formula $\dfrac{1}{1-u} = \displaystyle\sum_{n=0}^{\infty} u^n$ for $|u| < 1$ . The problems in this section develop facility with algebraic manipulation, center-shifting, and partial-fraction decomposition as tools for building power series representations.

Problem 4: Direct Geometric Series Substitution

Easy

Write the following function as a power series centered at $x = 0$ and state the interval of convergence:

f(x) = \frac{4}{1 + 2x^2}

Identify the substitution $u$ that puts $f(x)$ in the form $\dfrac{4}{1-u}$ .
Write out the power series and state the interval of convergence.

Hint

Rewrite

1 + 2x^2 = 1 - (-2x^2)

, so the geometric series formula applies with

u = -2x^2

. Convergence requires

|u| < 1

. Solve for

x

View Solution

Solution to part 1:

Write $f(x) = \dfrac{4}{1-(-2x^2)}$ . The substitution is $u = -2x^2$ .

Solution to part 2:

f(x) = 4\sum_{n=0}^{\infty}(-2x^2)^n = 4\sum_{n=0}^{\infty}(-2)^n x^{2n} = \sum_{n=0}^{\infty}(-1)^n \cdot 2^{n+2}\, x^{2n}.

Convergence requires $|-2x^2| < 1$ , i.e., $x^2 < \dfrac{1}{2}$ , so $|x| < \dfrac{1}{\sqrt{2}}$ . The interval of convergence is $\left(-\dfrac{1}{\sqrt{2}},\, \dfrac{1}{\sqrt{2}}\right)$ .

Problem 5: Center-Shifting and Non-Zero Center

Medium

Find a power series representation for the function centered at $a = 1$ :

f(x) = \frac{1}{x}

Rewrite $f(x)$ by expressing $x$ in terms of $(x - 1)$ so that a geometric series can be applied.
Write the resulting power series and state its interval of convergence.

Hint

Write

x = 1 - (1 - x) = 1 + (x - 1)

, so

\dfrac{1}{x} = \dfrac{1}{1+(x-1)} = \dfrac{1}{1-[-(x-1)]}

. Now apply the geometric series formula with

u = -(x-1)

View Solution

Solution to part 1:

\frac{1}{x} = \frac{1}{1-(-(x-1))} = \sum_{n=0}^{\infty}[-(x-1)]^n = \sum_{n=0}^{\infty}(-1)^n (x-1)^n.

Solution to part 2:

Convergence requires $|-(x-1)| = |x-1| < 1$ , i.e., $x \in (0, 2)$ . At $x = 0$ : $\sum (-1)^n(-1)^n = \sum 1$ , which diverges. At $x = 2$ : $\sum (-1)^n$ , which diverges. The interval of convergence is $(0,\, 2)$ .

Problem 6: Partial Fractions and Power Series

Hard

Express the following function as a single power series centered at $x = 0$ and determine its interval of convergence:

f(x) = \frac{3}{(1-x)(1+2x)}

Decompose $f(x)$ into partial fractions.
Write each fraction as a power series and combine into a single series with a general term formula.
State the interval of convergence for the combined series.

Hint

Write

\dfrac{3}{(1-x)(1+2x)} = \dfrac{A}{1-x} + \dfrac{B}{1+2x}

and solve for

A

and

B

. Each fraction yields a geometric series. The interval of convergence of the combined series is determined by the smaller of the two radii.

View Solution

Solution to part 1:

Set $3 = A(1+2x) + B(1-x)$ . At $x = 1$ : $3 = 3A \Rightarrow A = 1$ . At $x = -\tfrac{1}{2}$ : $3 = \tfrac{3}{2}B \Rightarrow B = 2$ .

f(x) = \frac{1}{1-x} + \frac{2}{1+2x}.

Solution to part 2:

\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, \quad |x| < 1.

\frac{2}{1+2x} = \frac{2}{1-(-2x)} = 2\sum_{n=0}^{\infty}(-2x)^n = \sum_{n=0}^{\infty}(-1)^n 2^{n+1} x^n, \quad |x| < \tfrac{1}{2}.

Combining:

f(x) = \sum_{n=0}^{\infty}\bigl[1 + (-1)^n 2^{n+1}\bigr]\,x^n.

Solution to part 3:

The first series converges for $|x| < 1$ and the second for $|x| < \dfrac{1}{2}$ . The combined series is valid on the intersection: $\left(-\dfrac{1}{2},\, \dfrac{1}{2}\right)$ .

Differentiation and Integration of Power Series

Within the interval of convergence, a power series can be differentiated and integrated term by term, as if it were a polynomial. This powerful property allows us to derive new series from known ones and to evaluate integrals that have no closed form in elementary functions. The following exercises build from straightforward differentiation to using integration to represent transcendental functions.

Problem 7: Differentiating a Known Power Series

Easy

Recall that $\dfrac{1}{1-x} = \displaystyle\sum_{n=0}^{\infty} x^n$ for $|x| < 1$ .

Differentiate both sides term by term to find a power series for $\dfrac{1}{(1-x)^2}$ .
Use the result from part 1 to write a power series for $\dfrac{x}{(1-x)^2}$ , and state the interval of convergence.

Hint

Differentiating

\sum x^n

term by term gives

\sum n x^{n-1}

. The derivative of

\dfrac{1}{1-x}

\dfrac{1}{(1-x)^2}

. For part 2, multiply both sides of the part-1 result by

x

View Solution

Solution to part 1:

\frac{d}{dx}\left(\sum_{n=0}^{\infty} x^n\right) = \sum_{n=1}^{\infty} n\,x^{n-1} = \frac{1}{(1-x)^2}, \quad |x| < 1.

Solution to part 2:

Multiply through by $x$ :

\frac{x}{(1-x)^2} = \sum_{n=1}^{\infty} n\,x^n = x + 2x^2 + 3x^3 + \cdots, \quad |x| < 1.

The interval of convergence is $(-1,\,1)$ (differentiation preserves the radius; endpoints must be rechecked, and both yield divergent series here).

Problem 8: Integrating to Find a Series for a Logarithm

Medium

Starting from the geometric series $\dfrac{1}{1+x} = \displaystyle\sum_{n=0}^{\infty}(-1)^n x^n$ for $|x| < 1$ :

Integrate both sides from $0$ to $x$ to obtain a power series for $\ln(1+x)$ .
Use the result to find the exact value of the series $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}$ .

Hint

Integrate

\displaystyle\int_0^x \frac{1}{1+t}\,dt = \ln(1+x)

and

\displaystyle\int_0^x \sum(-1)^n t^n\,dt

term by term. For part 2, set

x = 1

and check that

x = 1

is within the interval of convergence of the resulting series.

View Solution

Solution to part 1:

\ln(1+x) = \int_0^x \frac{1}{1+t}\,dt = \sum_{n=0}^{\infty}(-1)^n \int_0^x t^n\,dt = \sum_{n=0}^{\infty}(-1)^n \frac{x^{n+1}}{n+1} = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\,x^n.

The radius of convergence is 1. At $x = 1$ the series is the alternating harmonic series, which converges. At $x = -1$ it is $-\sum 1/n$ , which diverges. Hence the interval of convergence is $(-1,\,1]$ .

Solution to part 2:

Set $x = 1$ (which is in the interval of convergence):

\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} = \ln(1+1) = \ln 2.

Problem 9: Series Representation of a Definite Integral

Hard

The integral $\displaystyle\int_0^{0.5} e^{-x^2}\,dx$ cannot be expressed in closed form using elementary functions.

Write $e^{-x^2}$ as a power series centered at $x = 0$ by substituting into the Maclaurin series for $e^t$ .
Integrate the series term by term from $0$ to $0.5$ to obtain a numerical series for the integral.
Approximate the integral by summing the first four terms of the series and estimate the error using the alternating series estimation theorem.

Hint

Start from

e^t = \displaystyle\sum_{n=0}^{\infty}\dfrac{t^n}{n!}

and substitute

t = -x^2

. After integration, you get a series of the form

\displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n}{n!\,(2n+1)} (0.5)^{2n+1}

. This is an alternating series with decreasing terms; the error is bounded by the first omitted term.

View Solution

Solution to part 1:

e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots

This converges for all $x \in \mathbb{R}$ .

Solution to part 2:

\int_0^{0.5} e^{-x^2}\,dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_0^{0.5} x^{2n}\,dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!\,(2n+1)}\left(\frac{1}{2}\right)^{2n+1}.

Solution to part 3:

The first four terms ( $n = 0, 1, 2, 3$ ):

$n$	Term value
0	$\dfrac{1}{1\cdot 1}\cdot\dfrac{1}{2} = 0.500000$
1	$-\dfrac{1}{1\cdot 3}\cdot\dfrac{1}{8} = -0.041667$
2	$\dfrac{1}{2\cdot 5}\cdot\dfrac{1}{32} = 0.003125$
3	$-\dfrac{1}{6\cdot 7}\cdot\dfrac{1}{128} \approx -0.000186$

Partial sum: $0.500000 - 0.041667 + 0.003125 - 0.000186 \approx 0.461272$ .

The next term (the error bound) is $\dfrac{1}{24\cdot 9}\cdot\dfrac{1}{512} \approx 0.0000090$ . Thus:

\int_0^{0.5} e^{-x^2}\,dx \approx 0.4613 \quad \text{with error} < 0.000009.

Taylor and Maclaurin Series

Taylor and Maclaurin series allow us to represent a wide class of smooth functions as power series by computing successive derivatives at a center point. These series are indispensable in analysis, numerical methods, and physics. The following problems develop fluency with deriving Taylor series from scratch, using known Maclaurin series as building blocks, and determining the validity of such representations via Taylor’s Remainder Theorem.

Problem 10: Maclaurin Series from Derivatives

Easy

Find the Maclaurin series for $f(x) = \cos(x)$ directly from the definition.

Compute the first five derivatives of $f(x) = \cos x$ , evaluate them at $x = 0$ , and identify the pattern.
Write the Maclaurin series in sigma notation and state its interval of convergence.

Hint

The pattern of derivatives of

\cos x

cycles with period 4:

\cos x,\,-\sin x,\,-\cos x,\,\sin x,\,\cos x,\,\ldots

. Evaluate each at 0. Only even-indexed terms survive (since

\sin 0 = 0

View Solution

Solution to part 1:

$n$	$f^{(n)}(x)$	$f^{(n)}(0)$
0	$\cos x$	$1$
1	$-\sin x$	$0$
2	$-\cos x$	$-1$
3	$\sin x$	$0$
4	$\cos x$	$1$

The pattern repeats: $f^{(2k)}(0) = (-1)^k$ , and all odd-order terms vanish.

Solution to part 2:

\cos x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\,x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots

Since the Ratio Test yields $R = \infty$ , the interval of convergence is $(-\infty, +\infty)$ .

Problem 11: Taylor Series at a Non-Zero Center

Medium

Find the Taylor series for $f(x) = \ln x$ centered at $a = 1$ .

Compute $f^{(n)}(x)$ for general $n \geq 1$ and evaluate at $x = 1$ .
Write the Taylor series and determine the interval of convergence by testing endpoints.

Hint

Note that

f'(x) = 1/x

f''(x) = -1/x^2

f'''(x) = 2/x^3

, \ldots A general formula is

f^{(n)}(x) = \dfrac{(-1)^{n-1}(n-1)!}{x^n}

for

n \geq 1

. At

x = 1

this simplifies. The resulting series is related to the one in Problem 6.

View Solution

Solution to part 1:

For $n \geq 1$ : $f^{(n)}(x) = \dfrac{(-1)^{n-1}(n-1)!}{x^n}$ , so $f^{(n)}(1) = (-1)^{n-1}(n-1)!$ . Also $f(1) = \ln 1 = 0$ .

Solution to part 2:

\ln x = \sum_{n=1}^{\infty}\frac{f^{(n)}(1)}{n!}(x-1)^n = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n.

The Ratio Test gives $R = 1$ , so the open interval is $(0, 2)$ . At $x = 2$ : $\displaystyle\sum \frac{(-1)^{n-1}}{n} = \ln 2$ , which converges. At $x = 0$ : $\displaystyle\sum \frac{-1}{n}$ , which diverges. The interval of convergence is $(0,\, 2]$ .

Problem 12: Taylor’s Remainder and Approximation Error

Hard

Consider approximating $\sin(0.3)$ using the Maclaurin polynomial of degree 5.

Write the degree-5 Maclaurin polynomial $P_5(x)$ for $\sin x$ and compute $P_5(0.3)$ .
Use Taylor’s Remainder Theorem to find an upper bound for the error $|R_5(0.3)| = |\sin(0.3) - P_5(0.3)|$ .
Verify that the approximation is accurate to at least six decimal places.

Hint

Taylor’s Remainder Theorem states

|R_n(x)| \leq \dfrac{M}{(n+1)!}|x-a|^{n+1}

, where

M

is a bound on

|f^{(n+1)}|

between

a

and

x

. All derivatives of

\sin x

are bounded by 1 in absolute value. The next term after degree 5 is the

x^7

term.

View Solution

Solution to part 1:

P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} = x - \frac{x^3}{6} + \frac{x^5}{120}.

P_5(0.3) = 0.3 - \frac{(0.3)^3}{6} + \frac{(0.3)^5}{120} = 0.3 - \frac{0.027}{6} + \frac{0.00243}{120}

= 0.3 - 0.004500 + 0.0000202 = 0.2955202.

Solution to part 2:

The remainder after the degree-5 polynomial involves $f^{(6)}(x) = -\sin x$ or $f^{(7)}(x) = -\cos x$ . Since $|f^{(6)}(x)| \leq 1$ for all $x$ :

|R_5(0.3)| \leq \frac{1}{6!}(0.3)^6 = \frac{(0.3)^6}{720} = \frac{0.000729}{720} \approx 1.0 \times 10^{-6}.

Solution to part 3:

The error bound is less than $10^{-6}$ , confirming accuracy to at least six decimal places. (The true value $\sin(0.3) \approx 0.295520207$ agrees with $P_5(0.3) \approx 0.295520$ to six decimal places.)

Operations on Power Series and Advanced Applications

Beyond the standard toolkit, power series support a range of algebraic operations – addition, multiplication, composition, and substitution – that allow complex functions to be represented in series form. These techniques are essential for solving ordinary differential equations via series methods and for evaluating limits of indeterminate forms using series expansions.

Problem 13: Evaluating a Limit Using Power Series

Medium

Evaluate the following limit using power series (do not use L’Hôpital’s rule):

\lim_{x \to 0}\frac{1 - \cos x - \tfrac{x^2}{2}}{x^4}

Substitute the Maclaurin series for $\cos x$ and simplify the numerator.
Cancel the appropriate powers of $x$ and take the limit.

Hint

Write

\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots

and subtract term by term in the numerator. You should find the numerator equals a specific multiple of

x^4

plus higher-order terms.

View Solution

Solution to part 1:

1 - \cos x - \frac{x^2}{2} = 1 - \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720}+\cdots\right) - \frac{x^2}{2}

= 1 - 1 + \frac{x^2}{2} - \frac{x^4}{24} + \cdots - \frac{x^2}{2} = -\frac{x^4}{24} + O(x^6).

Solution to part 2:

\frac{1 - \cos x - \tfrac{x^2}{2}}{x^4} = \frac{-\tfrac{x^4}{24} + O(x^6)}{x^4} = -\frac{1}{24} + O(x^2) \to \boxed{-\frac{1}{24}} \quad \text{as } x\to 0.

Problem 14: Power Series Solution to a Differential Equation

Hard

Find a power series solution centered at $x = 0$ for the initial value problem:

y' = 2xy, \quad y(0) = 1.

Assume $y = \displaystyle\sum_{n=0}^{\infty} a_n x^n$ and substitute into the equation to find a recurrence relation for $a_n$ .
Solve the recurrence relation using $a_0 = 1$ and write out the first five non-zero terms of the solution.
Identify the closed-form function that this series represents.

Hint

After substituting, equate coefficients of

x^n

on both sides. You will find that

(n+1)a_{n+1} = 2a_{n-1}

, which generates only even-indexed (or odd-indexed) non-zero terms depending on

a_0

. With

a_0 = 1

, odd terms vanish. Compare the resulting series to a known Maclaurin series.

View Solution

Solution to part 1:

Substituting $y = \displaystyle\sum_{n=0}^{\infty} a_n x^n$ and $y' = \displaystyle\sum_{n=1}^{\infty} n\,a_n x^{n-1}$ :

\sum_{n=1}^{\infty} n\,a_n x^{n-1} = 2x \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} 2a_n x^{n+1} = \sum_{n=1}^{\infty} 2a_{n-1}\,x^n.

Re-indexing the left side (let $m = n-1$ ):

\sum_{m=0}^{\infty}(m+1)a_{m+1}\,x^m = \sum_{m=1}^{\infty}2a_{m-1}\,x^m.

Comparing coefficients: $a_1 = 0$ (constant term on right is 0), and for $m \geq 1$ : $(m+1)a_{m+1} = 2a_{m-1}$ .

Solution to part 2:

With $a_0 = 1$ and $a_1 = 0$ , the recurrence gives:

a_2 = \frac{2a_0}{2} = 1,\quad a_3 = \frac{2a_1}{3} = 0,\quad a_4 = \frac{2a_2}{4} = \frac{1}{2},\quad a_6 = \frac{2a_4}{6} = \frac{1}{6},\ldots

In general, $a_{2k} = \dfrac{1}{k!}$ and all odd-indexed coefficients are zero. The first five non-zero terms:

y = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \frac{x^8}{4!} + \cdots

Solution to part 3:

This is precisely the Maclaurin series for $e^{x^2}$ , since:

e^{x^2} = \sum_{k=0}^{\infty}\frac{(x^2)^k}{k!} = \sum_{k=0}^{\infty}\frac{x^{2k}}{k!}.

The closed-form solution is $\boxed{y = e^{x^2}}$ , which can be verified directly: $y' = 2x e^{x^2} = 2xy$ and $y(0) = 1$ . $\checkmark$