Find the sum sum from n=0 to infinity (x^(4n+1))/(4n+1)
Find the sum $\displaystyle\sum_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}$.
Explanation of Method
This picks out only the exponents $\equiv1\pmod4$ from the general odd-power series $\sum x^{2j+1}/(2j+1)$. Splitting that series by the parity of $j$ shows it equals $\operatorname{artanh}(x)+\arctan(x)$ for even $j$ plus a sign-flipped piece for odd $j$ that cancels — so averaging the two known odd-power series (artanh and arctan) isolates exactly the terms we want.
- Recall the two odd-power series $\operatorname{artanh}(x)$ and $\arctan(x)$.
- Add them and observe which terms survive.
- Solve for the desired sum.
Step 1 · The two odd-power series
$$ \operatorname{artanh}(x) = \sum_{j=0}^{\infty}\frac{x^{2j+1}}{2j+1}, \qquad \arctan(x) = \sum_{j=0}^{\infty}\frac{(-1)^jx^{2j+1}}{2j+1}. $$
Step 2 · Add and observe cancellation
$$ \operatorname{artanh}(x)+\arctan(x) = \sum_{j=0}^{\infty}\frac{1+(-1)^j}{2j+1}x^{2j+1}. $$ The coefficient $1+(-1)^j$ equals $2$ for even $j$ and $0$ for odd $j$. Writing $j=2k$ for the surviving even terms: $$ \operatorname{artanh}(x)+\arctan(x) = 2\sum_{k=0}^{\infty}\frac{x^{4k+1}}{4k+1}. $$
Step 3 · Solve for the sum
$$ \sum_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1} = \frac12\Bigl[\operatorname{artanh}(x)+\arctan(x)\Bigr] = \frac14\ln\frac{1+x}{1-x}+\frac12\arctan x. $$ Numerical check at $x=\frac12$: $\frac14\ln3+\frac12\arctan\frac12\approx0.2747+0.2318=0.5065$, matching the partial sum $0.5+\frac1{32\cdot5}+\cdots\approx0.506$.
Summary Table
| Property | Value |
|---|---|
| Sum | $\frac14\ln\frac{1+x}{1-x}+\frac12\arctan x$ |
| Equivalent form | $\frac12[\operatorname{artanh}(x)+\arctan(x)]$ |
| Region of validity | $|x|<1$ |
Additional Notes
- This "averaging trick" — adding a series and its sign-alternated twin to isolate every-other term — is a real-variable substitute for the roots-of-unity filter, and generalizes to extracting any arithmetic progression of exponents from a known series.
- The result combines Problem 272's $\ln\frac{1+x}{1-x}$ (giving $\operatorname{artanh}$) with Problem 267's $\arctan x$ series, both previously derived in this collection.
- Extracting exponents $\equiv3\pmod4$ instead would use the difference $\operatorname{artanh}(x)-\arctan(x)$ rather than the sum, since that combination cancels the even-$j$ terms instead.