`1/4+2/7+3/12+.....+n/(n^2+3)+........`
We can write the series as `sum_(n=1)^oon/(n^2+3)`
The 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=n/(n^2+3)`
Consider `f(x)=x/(x^2+3)`
Refer to the attached graph of the function. From the graph we observe that the function is positive, continuous and decreasing on...
`1/4+2/7+3/12+.....+n/(n^2+3)+........`
We can write the series as `sum_(n=1)^oon/(n^2+3)`
The 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=n/(n^2+3)`
Consider `f(x)=x/(x^2+3)`
Refer to the attached graph of the function. From the graph we observe that the function is positive, continuous and decreasing on the interval `[1,oo)`
We can also determine whether function is decreasing by finding the derivative f'(x) such that `f'(x)<0` for `x>=1`
Now let's determine whether the corresponding improper integral `int_1^oox/(x^2+3)dx` converges or diverges.
`int_1^oox/(x^2+3)dx=lim_(b->oo)int_1^bx/(x^2+3)dx`
Let's first evaluate the indefinite integral `intx/(x^2+3)dx`
Apply integral substitution:`u=x^2+3`
`=>du=2xdx`
`intx/(x^2+3)dx=int1/u(du)/2`
Take the constant out and use the common integral:`int1/xdx=ln|x|`
`=1/2ln|u|+C` where C is a constant
Substitute back `u=x^2+3`
`=1/2ln|x^2+3|+C`
`int_1^oox/(x^2+3)dx=lim_(b->oo)[1/2ln|x^2+3|]_1^b`
`=lim_(b->oo)1/2ln|b^2+3|-1/2ln|1^2+3|`
`=oo-1/2ln4`
`=oo`
Since the integral `int_1^oox/(x^2+3)dx` diverges, we can conclude from the integral test that the series diverges.
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