Question

Does `a_n=1/(n^2+1)` converge?

Answer 1

The sequence defined by `a_{n}=1/(n^2+1)` converges to zero. The corresponding infinite series `sum_{n=1}^{infty}1/(n^2+1)` converges to `(pi coth(pi)-1)/2 approx 1.077`.

Explanation:

That the sequence defined by `a_{n}=1/(n^2+1)` converges to zero is clear (if you wanted to be rigorous, for any `epsilon > 0`, the condition `0 < 1/(n^2+1) < epsilon` is equivalent to choosing `n` so that `n > sqrt(1/epsilon - 1)`, which, for any `0 < epsilon < 1` can definitely be done).

The series `sum_{n=1}^{infty}1/(n^2+1)` is most easily seen to converge by the comparison test. Let `b_{n}=1/n^{2}`, note that `0 leq a_{n} leq b_{n}` for all positive integers `n`, and note that `sum_{n=1}^{infty}b_{n}` converges since it's a `p`-series with `p=2 > 1` (the integral test can also be used to prove `sum_{n=1}^{infty}b_{n}` converges). Therefore, `sum_{n=1}^{infty}a_{n}=sum_{n=1}^{infty}1/(n^2+1)` converges by the comparison test.

To see that it converged to `(pi coth(pi)-1)/2 approx 1.077`, I used Wolfram Alpha with the input: sum (1/(k^2+1)) as k goes from one to infinity.

Note: `coth(x)=1/tanh(x)=cosh(x)/sinh(x)=(e^{x}+e^{-x})/(e^{x}-e^{-x})`