serie $\sum_{k=0}^{\infty} \left (\frac{1-k}{1+k}\right )^k$

Question

Can somebody please help me with this serie? $$\sum_{k=0}^{\infty} \left (\frac{1-k}{1+k}\right )^k $$ $$\lim_{k\to\infty} \frac{1-k}{1+k}=-1 $$ But i think that my solution is wrong (i did root test first) how should i evaluate this?

any suggestion would be great

Thanks

Answer 1

Write it as $(-1)^k (\frac{k-1}{k+1})^k = (-1)^k (1 - \frac2{k+1})^k$, and now it's easy to see that

• $(1 - \frac{2}{k+1})^k \to e^{-2}$ as $k \to \infty$;
• $(-1)^k (1 - \frac2{k+1})^k$ does not have a limit as $k \to \infty$: it alternates between numbers close to $e^{-2}$ and numbers close to $-e^{-2}$.

Therefore the sum cannot converge.