Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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! ;)

share|cite|improve this question
Haha now I dont :) – Roupam Ghosh Jun 21 '12 at 14:41
up vote 10 down vote accepted

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 $n$.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.