`sum_(n=1)^ooarctan(n)/(n^2+1)`
Integral test is applicable if f is positive, continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series `sum_(n=1)^ooa_n` converges or diverges if and only if the improper integral `int_1^oof(x)dx` converges or diverges.
For the given series `a_n=arctan(n)/(n^2+1)`
Consider `f(x)=arctan(x)/(x^2+1)`
Refer the attached graph of the function. From the graph, we observe that the function is positive , continuous and decreasing for `x>=1`
We an apply integral test as the function...
`sum_(n=1)^ooarctan(n)/(n^2+1)`
Integral test is applicable if f is positive, continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series `sum_(n=1)^ooa_n` converges or diverges if and only if the improper integral `int_1^oof(x)dx` converges or diverges.
For the given series `a_n=arctan(n)/(n^2+1)`
Consider `f(x)=arctan(x)/(x^2+1)`
Refer the attached graph of the function. From the graph, we observe that the function is positive , continuous and decreasing for `x>=1`
We an apply integral test as the function satisfies all the conditions for the integral test.
Now let's determine whether the corresponding improper integral `int_1^ooarctan(x)/(x^2+1)dx` converges or diverges.
`int_1^ooarctan(x)/(x^2+1)dx=lim_(b->oo)int_1^b arctan(x)/(x^2+1)dx`
Let's first evaluate the indefinite integral `intarctan(x)/(x^2+1)dx` ,
Apply integral substitution:`u=arctan(x)`
`du=1/(x^2+1)dx`
`=intudu`
Apply power rule,
`=u^2/2`
Substitute back `u=arctan(x)`
`=1/2(arctan(x))^2+C` where C is a constant
`int_1^ooarctan(x)/(x^2+1)dx=lim_(b->oo)[1/2(arctan(x))^2]_1^b`
`=lim_(b->oo)1/2[(arctan(b))^2-(arctan(1))^2]`
`=1/2[(pi/2)^2-(pi/4)^2]`
`=1/2(pi^2/4-pi^2/16)`
`=1/2((4pi^2-pi^2)/16)`
`=3/32pi^2`
Since the integral `int_1^ooarctan(x)/(x^2+1)dx` converges, we conclude from the integral test that the series converges.
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