Question

determine this whether is converge or diverge, if diverge, find its sum. (question going to be added now)

Answer 1

\[\sum_{n=1}^{\infty} 2/(n^2+4n+3) \]

Answer 2

sorry , if converge, find its sum

Answer 3

Well, it converges, and you can compare with 2/n^2, which is a p-series with p > 1. But calculating the sum precisely is another thing...

Answer 4

it converges by nth term test it goes to zero

Answer 5

we use the geometry rule?

Answer 6

for sum yes i guess we use geometric rule

Answer 7

What geometric rule? This doesn't look like a geometric series, does it?

Answer 8

sorry is series nor rule mistake typying

Answer 9

if u do show then show the procedure that was is asking in the texbook

Answer 10

i guess that is one way to find the sum if the result converge

Answer 11

If it's a geometric sequence, what is r here? I did calculate some terms and they seem to approach 5/6, but I am somewhat numbed, I guess. Can't see how to find the precise sum of that.

Answer 12

I mean, their sum seems to approach 5/6.

Answer 13

if opositive then is diverger when is infinity

Answer 14

thank so much for the help bmp

Answer 15

Was that right? :-O

Answer 16

the series can become :
\[\sum_{n=1}^{\infty} {(\frac{1}{n+1}-\frac{1}{n+3})}=\sum_{n=1}^{\infty} {\frac{1}{n+1}}-\sum_{n=1}^{\infty} {\frac{1}{n+3}}\]
We can see that
\[\sum_{n=1}^{\infty} {\frac{1}{n+1}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....\]
\[\sum_{n=1}^{\infty} {\frac{1}{n+3}}=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+....\]So when we use subtraction here for \[n\to \infty\]
Its sum is only \[\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\] as bmp says

Answer 17

Ahh, that makes sense. Kudos for seeing more than I did :-)