# Calculus 2 - Infinite Series

This is a similar question that I have to the other one that I posted.

Determine if the following series converges or diverges.

sum[k=1,inf] 3/(k(k+3))

I applied the limit comparison test with 1/k^2
also k(k+3))=k^2+3k

lim k->inf [ 3/(k^2+3k) ]/[ 1/k^2 ] = lim k->inf (3k^2/(k^2+3k) =H 3

because sum[k=1,inf] 1/k^2 is a convergent p-series than sum[k=1,inf] 3/(k(k+3)) must also converge by the limit comparison test.

I plugged this into wolfram alpha
sum[n=1,inf] of 3/(k(k+3))
and it said that the sum does not converge by the limit comparison test. Am I doing something wrong?

Thanks for any help.

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Dick
Homework Helper
Never trust Wolfram Alpha. Reagan said "trust but verify". Plug this into WA. "sum 3/(k*(k+3)) for k from 1 to infinity". You managed to confuse the poor thing somehow. Probably the [n=1,inf] part again.

Last edited:
Ray Vickson
Homework Helper
Dearly Missed
This is a similar question that I have to the other one that I posted.

Determine if the following series converges or diverges.

sum[k=1,inf] 3/(k(k+3))

I applied the limit comparison test with 1/k^2
also k(k+3))=k^2+3k

lim k->inf [ 3/(k^2+3k) ]/[ 1/k^2 ] = lim k->inf (3k^2/(k^2+3k) =H 3

because sum[k=1,inf] 1/k^2 is a convergent p-series than sum[k=1,inf] 3/(k(k+3)) must also converge by the limit comparison test.

I plugged this into wolfram alpha
sum[n=1,inf] of 3/(k(k+3))
and it said that the sum does not converge by the limit comparison test. Am I doing something wrong?

Thanks for any help.
Wolfram Alpha keeps giving you wrong answers (on this and some other problems). What would you conclude from that?

RGV

Never trust Wolfram Alpha. Reagan said "trust but verify". Plug this into WA. "sum 3/(k*(k+3)) for k from 1 to infinity". You managed to confuse the poor thing somehow. Probably the [n=1,inf] part again.
Oh I screwed up the indexes. Entering that actually gives me the right answer. Thanks. You can conclude that woflram alpha isn't the be all end all super calculator.