Question

Determine the sum of this series

Answer 1

\[\sum_{n=1}^{+\infty} {2n+1 \over n^{4} + 2n^{3} + n^{2} }\]

Answer 2

I have determined that the series is convergent using limit comparison, however, I am confused as to obtaining the sum.

Answer 3

Is the sum equivalent to the limit you find when comparing \[a _{n}\] to \[b_{n}\] ?

Answer 4

Have you tried decomposing the nth term into partial fractions?

Answer 5

No, but I will try that.

Answer 6

Thank you

Answer 7

Another question:
I am unclear about the syntax of series. In my initial series, what would be considered the nth term?

Answer 8

It would be \[{2n+1 \over n^{4} + 2n^{3} + n^{2} }\]

Answer 9

Then my question is this:
When you evaluate \[\lim_{n \rightarrow \infty}{2n+1 \over n^{4} + 2n^{3} + n^{2} }\], using the technique of pulling out the largest power (n^4) and evaluating the equation plugging infinity in for n, you get 1 as a result. Does this mean that the nth term goes to 1?

Answer 10

And if so, wouldn't that contradict the necessary condition for convergence of the nth term going to 0?

Answer 11

No, the limit that you have is indeed 0. You're probably talking about the nth partial sum going to 1.

Answer 12

If so, is the method I used valid for finding the nth partial sum?

Answer 13

\[\sum_{n=1}^{\infty} {\frac{2n+1}{n^2(n+1)^2}}=\sum_{n=1}^{\infty} {\frac{1}{n^2}}-\sum_{n=1}^{\infty} {\frac{1}{(n+1)^2}}\]
So this series is convergent.
Calculate the sum :
\[\sum_{n=1}^{\infty} {\frac{1}{n^2}}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...\]
\[\sum_{n=1}^{\infty} {\frac{1}{(n+1)^2}}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...\]
Subtraction here, the sum is 1 :)