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This is a piece of a much tougher infinite sum I'm trying to get. I think it should be a simple answer but having trouble knowing how to approach it. Thanks for the help!

Does this sum converge? $$\sum_{n=2}^{\infty} \frac{n}{n-1}$$

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Are you sure you wish to include n=1? –  Myself Apr 19 '11 at 1:09
    
thanks, will fix –  usul Apr 19 '11 at 1:10
    
Hint: It is worth writing down some of the first few terms of this sequence. For example, $2,\frac{3}{2},\frac{4}{3},\frac{5}{4},\frac{6}{5}\dots$. You should notice that all of the terms are $\geq 1$. –  Amitesh Datta Jun 12 '11 at 11:58

2 Answers 2

up vote 11 down vote accepted

The series fails the divergence test; the terms don't go to zero: $$\lim_{n\to\infty}\frac{n}{n-1} = 1\neq 0.$$ So this series does not converge.

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Oops, thanks. You know how you get wrapped up in something and need to take a step back and slow down? Yeah.... –  usul Apr 19 '11 at 1:12

First of all, you have to change the start of that sum, since for $n=1$ you'll be dividing by $0$.

Now, we can talk about the same series, but starting from $n=2$: $\sum_{n=2}^{\infty} \frac{n}{n-1}$

Let $(s_n)$ be the sequence that generates this series: $s_n=\frac{n}{n-1},n \geq 2$. Note that for every $n\geq 2,s_n \geq 1$, therefore: $$ \sum_{n=2}^{\infty} 1 \leq \sum_{n=2}^{\infty} \frac{n}{n-1} $$

But the LHS diverges, therefore our series also diverges.

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