Does the series sum[k=1,inf] 3/(k(k+3)) converge or diverge?

[GreenPrint]

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.

 

[Dick]

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.

 

[Ray Vickson]

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

 

[GreenPrint]

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.