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Question: Using the integral test, how do you show whether \(\sum{\dfrac{1}{{{n}^{2}}+1}}\)diverges or converg...

Using the integral test, how do you show whether ∑1n2+1\sum{\dfrac{1}{{{n}^{2}}+1}}diverges or converges from n=1n=1 to infinity?

Explanation

Solution

We will use the integral test on the term ∑1n2+1\sum{\dfrac{1}{{{n}^{2}}+1}}from n=1n=1 to infinity, the condition for the integral test is, given a function f(x)f(x), the condition for the function to converge is if the function is continuous, positive and decreasing on the interval [1,∞)[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,∞)[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 ∑1n2+1\sum{\dfrac{1}{{{n}^{2}}+1}} which is in the range [1,∞)[1,\infty ) therefore, it can be written in the summation format as ∑n=1∞1n2+1\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{n}^{2}}+1}}
Consider f(x)=1x2+1f(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 xx.
On differentiating, we get:
⇒f′(x)=ddx1x2+1\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:
⇒f′(x)=ddx(x2+1)−1\Rightarrow f'(x)=\dfrac{d}{dx}{{\left( {{x}^{2}}+1 \right)}^{-1}}
Now on differentiating and using chain rule, we get:
⇒f′(x)=−1(x2+1)−2ddx(x2+1)\Rightarrow f'(x)=-1{{\left( {{x}^{2}}+1 \right)}^{-2}}\dfrac{d}{dx}({{x}^{2}}+1)
On completing the derivative, we get:
⇒f′(x)=−(x2+1)−2×2x\Rightarrow f'(x)=-{{\left( {{x}^{2}}+1 \right)}^{-2}}\times 2x
On rearranging the terms, we get:
⇒f′(x)=−2x(x2+1)2\Rightarrow f'(x)=-\dfrac{2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}}
Now this function is negative for all the positive values of xx 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,∞)[1,\infty ).
We can write the equation as:
⇒∫1∞1x2+1\Rightarrow \int\limits_{1}^{\infty }{\dfrac{1}{{{x}^{2}}+1}}
We know that ⇒∫1x2+1=tan⁡−1(x)\Rightarrow \int{\dfrac{1}{{{x}^{2}}+1}}={{\tan }^{-1}}\left( x \right) therefore on integrating and taking the limits, we get:
⇒[tan⁡−1(x)]1∞\mathop{\Rightarrow [{{\tan }^{-1}}\left( x \right)]}_{1}^{\infty }
Now on simplifying, we get:
⇒tan⁡−1(∞)−tan⁡−1(1)\Rightarrow {{\tan }^{-1}}\left( \infty \right)-{{\tan }^{-1}}\left( 1 \right)
On taking the values on the inverse function, we get:
⇒π2−π4\Rightarrow \dfrac{\pi }{2}-\dfrac{\pi }{4}
On simplifying, we get:
⇒π4\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.