# [SOLVED] sums exist or do not exist

• Nov 21st 2009, 08:25 AM
SubZero
[SOLVED] sums exist or do not exist
Hi,can some one please help me with the following sums as i need to prove that they either exist or do not exist:

We do NOT say the sum exists if it is infinity.

1. $\sum _{k=1}^{\infty }{{k}^{-1}-{k}^{-2}}$

I know that:

this sequnce $\sum _{k=1}^{\infty }{k}^{-1}$ doesnot exist

but this sequence $\sum _{k=1}^{\infty }-{k}^{-2}$does exist

so i am confused about if this sequence $\sum _{k=1}^{\infty }{{k}^{-1}-{k}^{-2}}$ exists

2. i am not sure if the following sum exists

$\sum _{k=1}^{\infty }{{k}^{-1}- \left( k+1 \right) ^{-1}}$
• Nov 21st 2009, 08:47 AM
Jose27
The first one does not exist since if it did $\sum_{k=1}^{\infty } (k^{-1} + k^{-2}) -\sum_{k=1}^{\infty } k^{-2} = \sum_{k=1}^{ \infty } k^{-1}$ would be finite.

For the second one notice that $\sum_{k=1}^{n} k^{-1} -(k+1)^{-1} = (1 -1/2)+(1/2-1/3)+...+(1/n-1/(n+1))=1-1/(n+1)$
• Nov 21st 2009, 08:47 AM
Plato
Do you know the limit comparison test?
Compare $\frac{1}{n}-\frac{1}{n^2}$ with $\frac{1}{n}$.
• Nov 21st 2009, 09:03 AM
SubZero
Quote:

Originally Posted by Plato
Do you know the limit comparison test?
Compare $\frac{1}{n}-\frac{1}{n^2}$ with $\frac{1}{n}$.

we are not able to use the the limit comparison test, since my lecture has not proved it.

However we are able to use the comparison test and the absolute convergence, but i don't know if that will help.
• Nov 21st 2009, 09:04 AM
SubZero
THANKS FOR THE HELP Plato AND Jose27