# Thread: Convergence of a series.

1. ## Convergence of a series.

I am supposed to show that it converges or diverges. If it diverges, I need to further demonstrate whether it converges absolutely or conditionally.

Here is the problem I am having trouble with so far that I happened to get right (through the use of my logic but I don't know how to write it in math terms - I'll explain in detail as you read):

Problem: sum of &#x28;&#x28;n&#x5e;2 &#x2b; 1&#x29;&#x2f;&#x28;2n&#x5e;2 &#x2b; 1&#x29;&#x29;&#x5e;n from 1 to inf - Wolfram|Alpha

My logic is that the limit as n->inf of (n^2 + 1)/(2n^2 + 1) is 1/2 and (1/2)^n is a geometric series and it therefore should converge but this seems to be not good enough to me because of the fact that the first term for isntance is 2/3 and the second one is 5/9 which are somewhat away from 1/2 but then on the other hand the part of my logic that says it is still okay is that even the first term even though it's not near 1/2 is less than 1 which, in a geometric series, implies convergence.

Any input would be GREATLY appreciated!

2. Originally Posted by s3a
I am supposed to show that it converges or diverges. If it diverges, I need to further demonstrate whether it converges absolutely or conditionally.

Here is the problem I am having trouble with so far that I happened to get right (through the use of my logic but I don't know how to write it in math terms - I'll explain in detail as you read):

Problem: sum of &#x28;&#x28;n&#x5e;2 &#x2b; 1&#x29;&#x2f;&#x28;2n&#x5e;2 &#x2b; 1&#x29;&#x29;&#x5e;n from 1 to inf - Wolfram|Alpha

My logic is that the limit as n->inf of (n^2 + 1)/(2n^2 + 1) is 1/2 and (1/2)^n is a geometric series and it therefore should converge but this seems to be not good enough to me because of the fact that the first term for isntance is 2/3 and the second one is 5/9 which are somewhat away from 1/2 but then on the other hand the part of my logic that says it is still okay is that even the first term even though it's not near 1/2 is less than 1 which, in a geometric series, implies convergence.

Any input would be GREATLY appreciated!
Because the typical term in this series is an expression raised to the power n, you should use the root test.

$\lim_{n\to\infty}\sqrt[n]{a_n} = L$

If L < 1, the series converges.
If L > 1, the series diverges.
If L = 1, the test is inconclusive.

Good luck!!

3. Originally Posted by s3a
I am supposed to show that it converges or diverges. If it diverges, I need to further demonstrate whether it converges absolutely or conditionally.

Here is the problem I am having trouble with so far that I happened to get right (through the use of my logic but I don't know how to write it in math terms - I'll explain in detail as you read):

Problem: sum of &#x28;&#x28;n&#x5e;2 &#x2b; 1&#x29;&#x2f;&#x28;2n&#x5e;2 &#x2b; 1&#x29;&#x29;&#x5e;n from 1 to inf - Wolfram|Alpha

My logic is that the limit as n->inf of (n^2 + 1)/(2n^2 + 1) is 1/2 and (1/2)^n is a geometric series and it therefore should converge but this seems to be not good enough to me because of the fact that the first term for isntance is 2/3 and the second one is 5/9 which are somewhat away from 1/2 but then on the other hand the part of my logic that says it is still okay is that even the first term even though it's not near 1/2 is less than 1 which, in a geometric series, implies convergence.

Any input would be GREATLY appreciated!
As Wolfram's Alpha tells you, one would ordinarily use the root test to figure it out, but your idea isn't bad either. Since, as you write $q_n := \frac{n^2 + 1}{2n^2 + 1}\rightarrow \frac{1}{2}$, for any $\varepsilon > 0$, no matter how small, you can find an $n_0$ such that for all $n>n_0$ your $q_n$ is held in a "sandwich" $0\leq \left(\frac{1}{2}-\varepsilon\right)^n\leq q_n^n\leq \left(\frac{1}{2}+\varepsilon\right)^n$.
P.S: In fact, the argument is even simpler than that: since the terms of your series are all positive, it suffices to show that the partial sums are bounded from above, thus is suffices that there exists an $n_0$ such that for all $n>n_0$ we have that $0\leq q_n^n \leq \left(\frac{3}{4}\right)^n$.