# Ways to compute the limit of $\sum_{k=1}^\infty \frac{k^{n-1}}{(k+1)^n}$ as $n\to\infty$?

Consider the sum

$$S(n) = \frac{1^{n-1}}{2^n} + \frac{2^{n-1}}{3^n} + \frac{3^{n-1}}{4^n} + \cdots \infty = \sum_{k=1}^\infty \frac{k^{n-1}}{(k+1)^n}$$

How do I find the value of $\lim_{n\to\infty}S(n)$?

I am guessing it would be zero. But then again that's a guess! ;)

-
Haha now I dont :) –  Roupam Ghosh Jun 21 '12 at 14:41

Since $n$ is fixed, as $k \to \infty$, $$\left({k\over k+1}\right)^n \to 1.$$ As a result, $${k^{n-1}\over (k+1)^n} \sim {1\over k}\qquad {\rm as}\; k\to\infty$$ Since $n$ is fixed, the sum defined by $S(n)$ diverges for all $n$. By the limit comparison test, the sum $S(n)$ diverges for all $n$.

-
Ah yes LCT... Thanks :) –  Roupam Ghosh Jun 21 '12 at 14:41
If you want you can use just the comparison test: ${k^{n-1}\over (k+1)^n}> {k^{n-1}\over (2k)^n}=\frac{1}{2^n}\frac{1}{k}$ –  N. S. Jun 21 '12 at 15:56
@N.S. ha you posted this while I was typing up the same thing.. see below –  toypajme Jun 21 '12 at 16:09

Another way to see it:

for fixed $n$ and $N$, since $2k\geq k+1$ for all $k$

$$\sum_{k=1}^{N} \frac{k^{n-1}}{(k+1)^n}\geq \sum_{k=1}^{N}\frac{k^{n-1}}{(2k)^n} \geq \sum_{k=1}^{N}\frac{1}{2^{n}k}$$

Taking the limit as $N$ goes to infinity, for any $n$, it follows that the series $S(n)$ is divergent. Therefore the $\lim_{n\to\infty} S(n)$ does not exist.

-