Chapter 8.2, Problem 5E

To determine

The first 10 partial sums, the graph of the sequence of the terms and the sequence of the partial sums and determine if the series is convergent or divergent and hence find the sum of the series.

Answer to Problem 5E

The first 10 partial sums are: $0.44721$ , $1.15432$ , $1.98637$ , $2.88080$ , $3.80927$ , $4.75796$ , $5.71948$ , $6.68962$ , $7.66581$ and $8.64639$ .

The given series is divergent.

Explanation of Solution

Given: The given series is ${\displaystyle \sum _{n=1}^{\infty }\frac{n}{\sqrt{n^2+4}}}$ .

Concepts Used: Let ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ be any series. Then the $n^{th}$ partial sum of the series is given by-

  $s_n={\displaystyle \sum _{k=1}^n{a}_k=a_1+a_2+\mathrm{....}+a_n}$

If the sequence $\left\{s_n\right\}$ is convergent and ${\lim }_{n\to \infty }{s}_n=s$ as a real number, then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is convergent and the sum of the series is equal to s , i.e.-

  ${\displaystyle \sum _{n=1}^{\infty }{a}_n}=s$

If the sequence $\left\{s_n\right\}$ is divergent, then the series is also divergent.

Also, if $\underset{n\to \infty }{\lim }{a}_n$ doesn’t exist or if $\underset{n\to \infty }{\lim }{a}_n\ne 0$ , then also the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is divergent.

Calculations: The first 10 terms of the given sequence are as follows-

  $a_1=0.447214$

  $a_2=0.707107$

  $a_3=0.832050$

  $a_4=0.894427$

  $a_5=0.928477$

  $a_6=0.948683$

  $a_7=0.961524$

  $a_8=0.970143$

  $a_9=0.976187$

  $a_{10}=0.980581$

The first 10 partial sums are as follows-

  $s_1={\displaystyle \sum _{n=1}^1\frac{n}{\sqrt{n^2+4}}}=0.44721$

  $s_2={\displaystyle \sum _{n=1}^2\frac{n}{\sqrt{n^2+4}}}=1.15432$

  $s_3={\displaystyle \sum _{n=1}^3\frac{n}{\sqrt{n^2+4}}}=1.98637$

  $s_4={\displaystyle \sum _{n=1}^4\frac{n}{\sqrt{n^2+4}}}=2.88080$

  $s_5={\displaystyle \sum _{n=1}^5\frac{n}{\sqrt{n^2+4}}}=3.80927$

  $s_6={\displaystyle \sum _{n=1}^6\frac{n}{\sqrt{n^2+4}}}=4.75796$

  $s_7={\displaystyle \sum _{n=1}^7\frac{n}{\sqrt{n^2+4}}}=5.71948$

  $s_8={\displaystyle \sum _{n=1}^8\frac{n}{\sqrt{n^2+4}}}=6.68962$

  $s_9={\displaystyle \sum _{n=1}^9\frac{n}{\sqrt{n^2+4}}}=7.66581$

  $s_{10}={\displaystyle \sum _{n=1}^{10}\frac{n}{\sqrt{n^2+4}}}=8.64639$

Hence, the graph of the terms of the sequence of the partial sums of the sequence is given by-

  

Now,

  $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{n}{\sqrt{n^2+4}}$

  $=\underset{n\to \infty }{\lim }\frac{1}{\frac{\sqrt{n^2+4}}{n}}$

  $=\underset{n\to \infty }{\lim }\frac{1}{\sqrt{\frac{n^2+4}{n^2}}}$

  $=\underset{n\to \infty }{\lim }\frac{1}{\sqrt{1+\frac{4}{n^2}}}$

  $=1$ (Since $\underset{n\to \infty }{\lim }\frac{4}{n^2}=0$ )

Since, $\underset{n\to \infty }{\lim }{a}_n\ne 0$ , thus, the given series is divergent.