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? $\sum_{n=2}^{\infty} \frac{n}{n-1}$
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? $\sum_{n=2}^{\infty} \frac{n}{n-1}$
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.
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.