Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 7

1/2+1/5+1/10+1/17+1/26+............
The series can be written as,
1/(1^2+1)+1/(2^2+1)+1/(3^2+1)+1/(4^2+1)+1/(5^2+1)+..........
So based on the above pattern we can write the series as,
sum_(n=1)^oo1/(n^2+1)
The integral test is applicable if f is positive, continuous and decreasing function on the infinite interval [k,oo) where k>=1 and a_n=f(x) . Then the series converges or diverges if and only if the improper integral int_k^oof(x)dx converges or diverges.
For the given series a_n=1/(n^2+1)
Consider f(x)=1/(x^2+1)

Graph of the function is attached. From the graph we can see that the function is positive, continuous and decreasing on the interval [1,oo)
Since the function satisfies the conditions for the integral test, we can apply integral test.
Now let's determine whether the corresponding improper integral int_1^oo1/(x^2+1)dx converges or diverges.
int_1^oo1/(x^2+1)dx=lim_(b->oo)int_1^b1/(x^2+1)dx
Use the common integral: int1/(x^2+1)dx=arctan(x)
=lim_(b->oo)[arctan(x)]_1^b
=lim_(b->oo)[arctan(b)-arctan(1)]
=lim_(b->oo)arctan(b)-arctan(1)
=pi/2-pi/4
=pi/4
Since the integral int_1^oo1/(x^2+1)dx converges, we conclude from the integral test that the series also converges.