Question

Using the integral test, how do you show whether $\sum{\dfrac{1}{{{n}^{2}}+1}}$diverges or converges from $n=1$ to infinity?

Answer 1

Hint: We will use the integral test on the term $\sum{\dfrac{1}{{{n}^{2}}+1}}$from $n=1$ to infinity, the condition for the integral test is, given a function $f(x)$, the condition for the function to converge is if the function is continuous, positive and decreasing on the interval $[1,\infty )$. We will first check whether the function decreases by taking the derivative of the function and then integrate the function in the limit $[1,\infty )$ and see if the function is positive in the domain. If both the conditions are fulfilled, we can conclude that the series is convergent.

Complete step-by-step answer:
We have the function as $\sum{\dfrac{1}{{{n}^{2}}+1}}$ which is in the range $[1,\infty )$ therefore, it can be written in the summation format as $\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{n}^{2}}+1}}$
Consider $f(x)=\dfrac{1}{{{x}^{2}}+1}$
Now to check whether the function is decreasing, we will differentiate the function and test for all the positive values of $x$.
On differentiating, we get:
$\Rightarrow f'(x)=\dfrac{d}{dx}\dfrac{1}{{{x}^{2}}+1}$
On taking the reciprocal of the function, we can write it in the exponent form as:
$\Rightarrow f'(x)=\dfrac{d}{dx}{{\left( {{x}^{2}}+1 \right)}^{-1}}$
Now on differentiating and using chain rule, we get:
$\Rightarrow f'(x)=-1{{\left( {{x}^{2}}+1 \right)}^{-2}}\dfrac{d}{dx}({{x}^{2}}+1)$
On completing the derivative, we get:
$\Rightarrow f'(x)=-{{\left( {{x}^{2}}+1 \right)}^{-2}}\times 2x$
On rearranging the terms, we get:
$\Rightarrow f'(x)=-\dfrac{2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}$
Now this function is negative for all the positive values of $x$ therefore, the series is decreasing.
Now we will check whether the function is always positive therefore, we will integrate the function in the limit $[1,\infty )$.
We can write the equation as:
$\Rightarrow \int\limits_{1}^{\infty }{\dfrac{1}{{{x}^{2}}+1}}$
We know that $\Rightarrow \int{\dfrac{1}{{{x}^{2}}+1}}={{\tan }^{-1}}\left( x \right)$ therefore on integrating and taking the limits, we get:
$\mathop{\Rightarrow [{{\tan }^{-1}}\left( x \right)]}_{1}^{\infty }$
Now on simplifying, we get:
$\Rightarrow {{\tan }^{-1}}\left( \infty \right)-{{\tan }^{-1}}\left( 1 \right)$
On taking the values on the inverse function, we get:
$\Rightarrow \dfrac{\pi }{2}-\dfrac{\pi }{4}$
On simplifying, we get:
$\Rightarrow \dfrac{\pi }{4}$, which is positive therefore, both the conditions are fulfilled which implies that the series is convergent.

Note: For a series to be conditionally convergent the sum of the positive terms diverges to positive infinity or sum of negative terms lead to negative infinity. For a series to be absolutely convergent the summation of the absolute value of all the summands is a finite term. The common mistake done in questions of series is that the denominator of the fraction should never be zero because it will result in a fallacy.