Vtyjkyuk

Show that
$$ sum_n=1^infty left( frac1n(n+1) right) = frac12+frac16+frac112+ ;... $$
converges and find its sum.

My solution so far:

I am thinking about finding the partial sum first and show that the series converges since its finite partial sum converges.

Now

$$ S_N=sum_n=1^N left( frac1n(n+1) right)=sum_n=1^N left( frac1n-frac1n+1 right)=frac12+frac16+frac112+;...= left( frac11-frac12 right)+left( frac12-frac13 right) + ;...+ left( frac1N-frac1N+1 right)$$

but I don't know how to go on with this. Now $ lim_N toinfty left( frac1N-frac1N+1 right)=0$ but the right answer should be $1$.

You're right that in the end $$lim_Nrightarrowinftyleft(frac1N-frac1N+1right)=0$$ But this is not the sum itself, for this, note that $$left(frac11-frac12right)+left(frac12-frac13right)+left(frac13-frac14right)+dots\=1+left(-frac12+frac12right)+left(-frac13+frac13right)+dots$$

You can use the Cauchy condensation test to show convergence:

$$sum_n=1^infty frac2^n2^n(2^n+1) = sum_n=1^infty frac12^n+1 le sum_n=1^infty frac12^n = 1$$

because the latter is a geometric series.

Hence $displaystylesum_n=1^infty frac1n(n+1)$ also converges.

See the first two brackets. $-frac12$ in the first bracket and $frac12$ in the second bracket cancel each other out. This canceling happens all through out.

The partial sum must be:
$$S_N=sum_n=1^N left( frac1n(n+1) right)=sum_n=1^N left( frac1n-frac1n+1 right)=\
frac12+frac16+frac112+;...colorredfrac1N-frac1N+1= \
left( colorbluefrac11-requirecancelcancelfrac12 right)+left( cancelfrac12-cancelfrac13 right) + ;...+ left( cancelfrac1N-colorbluefrac1N+1 right)=\
colorblue1-colorbluefrac1N+1$$
Hence:
$$lim_Ntoinfty S_N=lim_Ntoinfty left(1-frac1N+1right)=1.$$