1. ## Limit of series

I'm just starting to study limits of sequences and have been asked to determine if the series;

Sum (n=1 --> n = inf) 1/((n^2)+1) converges or diverges,

Is it correct for me to do the following,

Series will converge if the sequence (1/((n^2)+1) converges, hence;

take lim (n-->inf) (1/((n^2)+1)

and the limit = 0 as 1/inf is 0 in the limit.

?? is this correct? and should i be able to calculate the actual limit.

Cheers.

2. Originally Posted by monster
I'm just starting to study limits of sequences and have been asked to determine if the series;

Sum (n=1 --> n = inf) 1/((n^2)+1) converges or diverges,

Is it correct for me to do the following,

Series will converge if the sequence (1/((n^2)+1) converges, hence;

take lim (n-->inf) (1/((n^2)+1)

and the limit = 0 as 1/inf is 0 in the limit.

?? is this correct? and should i be able to calculate the actual limit.

Cheers.
Use comparsion test : $\displaystyle \frac{1}{n^2 + 1} < \frac{1}{n^2}$

BTW , the sum is $\displaystyle \frac{ \pi \coth(\pi) - 1}{2}$

3. Originally Posted by monster
I'm just starting to study limits of sequences and have been asked to determine if the series;

Sum (n=1 --> n = inf) 1/((n^2)+1) converges or diverges,

Is it correct for me to do the following,

Series will converge if the sequence (1/((n^2)+1) converges, hence;
You seem to be saying that the series converges if and only if the sequence converges but that not true. If the sequence $\displaystyle a_n$ converges to any number other than 0, the series, $\displaystyle \sum a_n$ does NOT converge. If the sequence converges to 0, the series may converge but not necessarily. for example 1/n converges to 0 but the series $\displaystyle \sum 1/n$ does NOT converge.

[/quote]take lim (n-->inf) (1/((n^2)+1)

and the limit = 0 as 1/inf is 0 in the limit.

?? is this correct? and should i be able to calculate the actual limit.

Cheers.[/QUOTE]
No, that is not correct. As simple pendulum says, you can use the comparison test to show that this particular series does converge. You could also use the integral test: $\displaystyle \int_0^\infty \frac{1}{x^2+ 1}= \left[arctan(x)]_0^\infty= \frac{\pi}{2}$ is finite so this series converges. There are, in general, no good ways to actually find the sum.