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})#