Find the sum sum from n=1 to infinity

Find the sum $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n2^{2n-1}x^{2n-1}}{2n-1}$.

Step 1 · Rewrite in terms of $2x$

$$ \frac{(-1)^n2^{2n-1}x^{2n-1}}{2n-1} = \frac{(-1)^n(2x)^{2n-1}}{2n-1}. $$

Step 2 · Reindex against $\arctan$

With $t=2x$, let $m=n-1$: $$ \sum_{n=1}^{\infty}\frac{(-1)^nt^{2n-1}}{2n-1} = \sum_{m=0}^{\infty}\frac{(-1)^{m+1}t^{2m+1}}{2m+1} = -\sum_{m=0}^{\infty}\frac{(-1)^mt^{2m+1}}{2m+1} = -\arctan t = -\arctan(2x). $$

Step 3 · Verify

First terms: $n=1$: $-2x$; $n=2$: $\frac{8x^3}3$; $n=3$: $-\frac{32x^5}5$. Expanding $-\arctan(2x)=-\left[2x-\frac{(2x)^3}3+\frac{(2x)^5}5-\cdots\right]=-2x+\frac{8x^3}3-\frac{32x^5}5+\cdots$ — matches term by term.

Summary Table

Additional Notes

• This is simply the $\arctan$ series (Problem 267) with $x$ rescaled to $2x$ and the overall sign flipped — the reindexing shift ($n\to n-1$) is what introduces the extra minus sign.
• Because the underlying series requires $|2x|\le1$, the region of convergence shrinks to $|x|\le\frac12$, half the radius of the original $\arctan$ series.
• This kind of rescaling appears throughout the collection whenever a series is presented with a "disguised" argument like $2x$ hidden inside powers and coefficients — always look for the substitution that reveals the standard series underneath.