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

Dec 10, 2015

The sequence defined by ${a}_{n} = \frac{1}{{n}^{2} + 1}$ converges to zero. The corresponding infinite series ${\sum}_{n = 1}^{\infty} \frac{1}{{n}^{2} + 1}$ converges to $\frac{\pi \coth \left(\pi\right) - 1}{2} \approx 1.077$.

#### Explanation:

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

The series ${\sum}_{n = 1}^{\infty} \frac{1}{{n}^{2} + 1}$ is most easily seen to converge by the comparison test. Let ${b}_{n} = \frac{1}{n} ^ \left\{2\right\}$, note that $0 \le q {a}_{n} \le q {b}_{n}$ 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, ${\sum}_{n = 1}^{\infty} {a}_{n} = {\sum}_{n = 1}^{\infty} \frac{1}{{n}^{2} + 1}$ converges by the comparison test.

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

Note: $\coth \left(x\right) = \frac{1}{\tanh} \left(x\right) = \cosh \frac{x}{\sinh} \left(x\right) = \frac{{e}^{x} + {e}^{- x}}{{e}^{x} - {e}^{- x}}$