`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...

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`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.