Exercises — Term-by-term differentiation and integration of power series

WHAT we do: apply the derivative rule — multiply the $n$-th coefficient by $n$, drop the power by one, and start the sum at $n=1$ (the $n=0$ term is a constant, its derivative is $0$). Here $a_n=\frac{1}{n+1}$, so $n{a}_n=\frac{n}{n+1}$: $f^{\prime }(x)={\sum }_{n=1}^{\infty }\frac{n}{n+1}{x}^{n-1}.$ WHY $R$ is unchanged: the original series has $R=1$ (it looks like the geometric series slowed by $\frac{1}{n+1}$), and differentiation never changes $R$. So the answer is valid on $|x|<1$.

WHAT we do: integrate each brick. Rule: divide the coefficient of $x^k$ by $(k+1)$ and raise the power to $k+1$. Here the power is $k=2n$, so $\int x^{2n}dx=\frac{x^{2n+1}}{2n+1}$: ${\int }_0^{x}g(t)dt={\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{2n+1}{x}^{2n+1}.$ WHAT it is: this is exactly the $\arctan x$ series (see Example 2 in the parent). The constant is $0$ because the definite integral from $0$ is $0$ at $x=0$.

WHY differentiate? Notice $\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x{)}^2}$ (chain rule on the left). So differentiating the known series hands us the unknown one for free. $\frac{1}{(1-x{)}^2}=\frac{d}{dx}{\sum }_{n=0}^{\infty }{x}^n={\sum }_{n=1}^{\infty }n{x}^{n-1}={\sum }_{m=0}^{\infty }(m+1)x^m.$ (Re-index $m=n-1$.) So $h(x)=1+2x+3x^2+4x^3+\cdots$, valid on $|x|<1$. Check the first two coefficients: $h(0)=1$ ✓, and the coefficient of $x$ is $2$.

WHY integrate? Because $\frac{d}{dx}\ln (1-x)=\frac{-1}{1-x}$. So $\ln (1-x)$ is an antiderivative of $-\frac{1}{1-x}=-\sum x^n$. Integrate term by term from $0$ to $x$: $\ln (1-x)=-{\sum }_{n=0}^{\infty }\frac{x^{n+1}}{n+1}=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots \right).$ Constant: $\ln (1-0)=\ln 1=0$, so $C=0$. Radius: integration keeps $R=1$.

At $x=1$: the series becomes $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$. This is an alternating series whose terms $\frac{1}{2n+1}$ decrease to $0$, so by the alternating-series test it converges (to $\frac{\pi }{4}$). At $x=-1$: plug in — odd powers flip sign, giving $-1+\frac{1}{3}-\frac{1}{5}+\cdots =-(1-\frac{1}{3}+\frac{1}{5}-\cdots )$, which also converges (to $-\frac{\pi }{4}$). Gained or lost? The original series $\sum (-1{)}^n{x}^{2n}$ at $x=\pm 1$ is $1-1+1-1+\cdots$, which diverges (terms don't go to $0$). After integrating, both endpoints now converge — so integration gained convergence at the endpoints. This matches the parent note's rule: $R$ is unchanged, but endpoints can improve when you integrate. See Abel's theorem (endpoint behaviour) for why the sum there really equals $\arctan (\pm 1)$.

Differentiate: $f^{\prime }(x)={\sum }_{n=1}^{\infty }\frac{n{x}^{n-1}}{n}={\sum }_{n=1}^{\infty }{x}^{n-1}={\sum }_{m=0}^{\infty }{x}^m=\frac{1}{1-x}$, radius still $1$. Endpoint $x=-1$: the derivative series is $1-1+1-1+\cdots$, which diverges (terms don't tend to $0$). Conclusion: the original converged at $x=-1$, the derivative does not. So differentiation lost convergence at that endpoint — the mirror image of L3.1. This is exactly the "DIRT: Test endpoints again" warning.

WHY start from L2.1? We already know ${\sum }_{n=1}^{\infty }n{x}^{n-1}=\frac{1}{(1-x{)}^2}$. Our target has one extra factor of $x$, so just multiply both sides by $x$: $S(x)={\sum }_{n=1}^{\infty }n{x}^n=x\cdot \frac{1}{(1-x{)}^2}=\frac{x}{(1-x{)}^2}.$ Numeric evaluation: put $x=\frac{1}{2}$ (which is inside $|x|<1$): ${\sum }_{n=1}^{\infty }\frac{n}{2^n}=\frac{1/2}{(1-1/2{)}^2}=\frac{1/2}{1/4}=2.$

WHY differentiate? The coefficient $\frac{1}{2n+1}$ signals an integral was done — so differentiating should simplify it. Differentiate term by term: $T^{\prime }(x)={\sum }_{n=0}^{\infty }\frac{(2n+1)x^{2n}}{2n+1}={\sum }_{n=0}^{\infty }{x}^{2n}=\frac{1}{1-x^2}(|x|<1).$ Recover $T$ by integrating back. Use the partial-fraction split $\frac{1}{1-x^2}=\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)$: $T(x)={\int }_0^x\frac{dt}{1-t^2}=\frac{1}{2}(-\ln (1-x)+\ln (1+x))=\frac{1}{2}\ln \frac{1+x}{1-x}.$ Constant: $T(0)=0$ and $\frac{1}{2}\ln 1=0$ ✓. This is the artanh (inverse hyperbolic tangent) series. Spot check at $x=\frac{1}{2}$: $T(\frac{1}{2})=\frac{1}{2}\ln 3\approx 0.5493$.

WHY a power series? Because term-by-term differentiation turns a differential equation into plain algebra on the coefficients (this is the method of Solving ODEs with power series). Step 1 — differentiate the ansatz. $y^{\prime }={\sum }_{n=1}^{\infty }n{a}_n{x}^{n-1}={\sum }_{m=0}^{\infty }(m+1)a_{m+1}{x}^m$ (re-index $m=n-1$). Step 2 — match to $y$. The equation $y^{\prime }=y$ means, coefficient by coefficient of $x^m$: $(m+1)a_{m+1}=a_m⟹a_{m+1}=\frac{a_m}{m+1}.$ Step 3 — use the initial condition. $y(0)=a_0=1$. Then $a_1=\frac{1}{1}=1$, $a_2=\frac{1}{2}$, $a_3=\frac{1}{2\cdot 3}=\frac{1}{6}$, and in general $a_n=\frac{1}{n!}.$ Step 4 — read off the function. $y={\sum }_{n=0}^{\infty }\frac{x^n}{n!}=e^x$, with radius $R=\infty$ (the ratio $\frac{a_{n+1}}{a_n}=\frac{1}{n+1}\to 0$, so it converges for every $x$).

Step 1 — integrate. From $\frac{1}{1+t^2}=\sum (-1{)}^n{t}^{2n}$ ($|t|<1$), integrate from $0$ to $x$: $\arctan x={\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{2n+1}{x}^{2n+1},|x|<1.$ Step 2 — endpoint convergence. At $x=1$ the series is $1-\frac{1}{3}+\frac{1}{5}-\cdots$, alternating with terms decreasing to $0$, so it converges (L3.1 showed this). Step 3 — why the sum equals $\arctan 1$. Convergence at the endpoint is not enough on its own; we invoke Abel's theorem (endpoint behaviour): if a power series converges at an endpoint, its sum there equals the limit of the function from inside. Since $\arctan$ is continuous and $\arctan 1=\frac{\pi }{4}$, ${\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{2n+1}=\arctan 1=\frac{\pi }{4}.$ That is the Leibniz formula. $■$

Term-by-term derivative of $\sum \frac{x^n}{n!}$ ::: $\sum \frac{x^n}{n!}$ again (it is $e^x$). Closed form of ${\sum }_{n\ge 1}n{x}^n$ ::: $\frac{x}{(1-x{)}^2}$. Value of ${\sum }_{n\ge 1}n/2^n$ ::: $2$. Closed form of ${\sum }_{n\ge 0}\frac{x^{2n+1}}{2n+1}$ ::: $\frac{1}{2}\ln \frac{1+x}{1-x}$. Theorem that pushes $\arctan$ series to $x=1$ ::: Abel's theorem.