Question

In this task, determine the convergence or divergence of the series.

n=1nn2+1\displaystyle\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^2+1}}

Solution

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Answered 1 year ago
Answered 1 year ago

Let n=1an=n=1nn2+1\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\dfrac{n}{\sqrt{n^2+1}}. Then consider

limnan=limnnn2+1=limn11+1/n2(Dividing both numerator and denominator by n)=11+0=1\begin{aligned}\lim_{n\rightarrow \infty}a_n&=\lim_{n\rightarrow \infty}\dfrac{n}{\sqrt{n^2+1}}\\&=\lim_{n\rightarrow \infty}\dfrac{1}{\sqrt{1+1/n^2}}&&(\text{Dividing both numerator and denominator by }n)\\&=\dfrac{1}{\sqrt{1+0}}\\&=1\end{aligned}

Thus we get limnan0\lim_{n\rightarrow \infty}a_n\neq 0 and then using nthn^{th} term test for divergence we get n=1an\sum_{n=1}^{\infty}a_n diverges.

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