Question

Find whether the following series are convergent or divergent:
${\displaystyle \frac{a+x}{1}}+{\displaystyle \frac{{(a+2x)}^2}{|\underset{―}{2}}}+{\displaystyle \frac{{(a+3x)}^3}{|\underset{―}{3}}}+......$.

Answer 1

Hint: To solve the question given above, we will first find out what are convergent and divergent series. Then we will find out the $n^{th}$ and ${(n+1)}^{th}$ term of the series and then we will determine the limit: ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{{(n+1)}^{th}term}{n^{th}term}}\right|}$. If the value of this limit will be less than 1, then the series will be convergent, otherwise it will be divergent.

Complete step by step answer:
Before we solve the question given above, we must know what is a convergent and what is a divergent series. A convergent series is that kind of series in which the terms of series appear to be approaching a particular number or the series appear to be converge. A divergent series is that kind of series in which the terms of the series appear to be moving away from O. Now, it is given in question that we have to decide whether the series is convergent or divergent. For this, we will find out the $n^{th}$ and ${(n+1)}^{th}$ term of the series.
We can see that every term of the series resembles the form ${\displaystyle \frac{{(a+rx)}^n}{|\underset{―}{r}}}$, $0<r<\mathrm{\infty }$. We know that the symbol $|\underset{―}{}$ represents the factorial of a positive number. Factorial can also be represented as ${}^{\prime }{!}^{\prime }$.
So, we get the $n^{th}$ term of the series is: ${\displaystyle \frac{{(a+nx)}^n}{|\underset{―}{n}}}={\displaystyle \frac{{(a+nx)}^n}{n!}}$.
And the ${(n+1)}^{th}$ term of the series is:${\displaystyle \frac{{[a+(n+1)x]}^{n+1}}{(n+1)!}}$ .
Now, we will find the value of the following limit:
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{{(n+1)}^{th}term}{n^{th}term}}\right|}$.
If the value of the above limit will be less than 1 then the series will be convergent, otherwise it will be divergent series,
$={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{{\displaystyle \frac{{[a+(n+1)x]}^{n+1}}{(n+1)!}}}{{\displaystyle \frac{{[a+nx]}^n}{n!}}}}\right|}$.
$={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{n!\times {[a+(n+1)x]}^{n+1}}{(n+1)!\times {[a+nx]}^n}}\right|}$.
$={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{n!}{(n+1)!}}\right|\left|{\displaystyle \frac{{[a+(n+1)x]}^{n+1}}{{[a+nx]}^n}}\right|}$.
Now, we will use the formula: $x!=x\times (x-1)!$. Thus, we will get:
$={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{n!}{(n+1)n!}}\right|\left|{\displaystyle \frac{{[a+(n+1)x]}^{n+1}}{{[a+nx]}^n}}\right|}$.
$={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{1}{n+1}}\right|\left|{\displaystyle \frac{{[a+(n+1)x]}^{n+1}}{{[a+nx]}^n}}\right|}$.
Now, here we will use an exponential identity as shown:
$a^{x+y}=a^x\times a^y$.
Thus, we will get:
$={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{a+[n+1]x}{n+1}}\right|\left|{\left[{\displaystyle \frac{a+(n+1)x}{a+nx}}\right]}^n\right|}$.
$={\displaystyle \underset{n\to \mathrm{\infty }}{lim}|{\displaystyle \frac{a}{n+1}}+x|\left|{\left[{\displaystyle \frac{a+nx+x}{a+nx}}\right]}^n\right|}$.
$={\displaystyle \underset{n\to \mathrm{\infty }}{lim}|x+{\displaystyle \frac{a}{n+1}}|\left|{(1+{\displaystyle \frac{x}{a+nx}})}^n\right|}$.
$=\left[{\displaystyle \underset{n\to \mathrm{\infty }}{lim}(x+{\displaystyle \frac{a}{n+1}})}\right]\times \left[{\displaystyle \underset{n\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{x}{a+nx}})}^n}\right].......\left(1\right)$.
Now, we assume that ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}(x+{\displaystyle \frac{a}{n+1}})=p}$ and ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{x}{a+nx}})}^n=g}$. We will calculate the values of p and g separately. Thus, we have:
$p={\displaystyle \underset{n\to \mathrm{\infty }}{lim}(x+{\displaystyle \frac{a}{n+1}})}$.
$\Rightarrow p={\displaystyle \underset{n\to \mathrm{\infty }}{lim}x+{\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{a}{n+1}}}}$.
$\Rightarrow p=x+{\displaystyle \frac{a}{\mathrm{\infty }}}$.
$\Rightarrow p=x+0$ .
$\Rightarrow p=x$.
Now, we will calculate q. Thus, we have:
$q={\displaystyle \underset{n\to \mathrm{\infty }}{lim}{(1+{\displaystyle \frac{x}{a+nx}})}^n}$.
We will take logarithm on both sides
$\log q={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\log {(1+{\displaystyle \frac{x}{a+nx}})}^n}$.
Now, we will apply another identity
$\log r^s=s\log r$.
$\log q={\displaystyle \underset{n\to \mathrm{\infty }}{lim}[n\log (1+{\displaystyle \frac{x}{a+nx}})]}$.
Now, we know that $\log (1+n)=x-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^3}{3}}......$.
$\log q={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left[n\left\{({\displaystyle \frac{x}{a+nx}}-{\displaystyle \frac{{\left({\displaystyle \frac{x}{a+nx}}\right)}^2}{2}}+..........)\right\}\right]}$.
$\log q={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{nx}{a+nx}}\right]}$.
In the above equation, we have neglected the terms with higher power because they will be zero when we will put $n=\mathrm{\infty }$. Thus, we will get:
x$\log q={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{a+nx-a}{a+nx}}\right]}$.
$\Rightarrow \log q={\displaystyle \underset{n\to \mathrm{\infty }}{lim}[1-{\displaystyle \frac{a}{a+nx}}]}$.
$\Rightarrow \log q=1$.
$\Rightarrow q=e$.
Now, we will substitute the values of p and q in equation (1)
$=\left|x\right|e$.
Now for convergent series, $\left|x\right|e<1\Rightarrow \left|x\right|<{\displaystyle \frac{1}{e}}\Rightarrow x<{\displaystyle \frac{1}{e}}$

For $x\ge {\displaystyle \frac{1}{e}}$, the series will be divergent.

Note: We have used the expansion of $\log (1+x)$ while calculating the value of q. We cannot use this expansion for any value of x. The expansion can be used only when $\left|x\right|<1$. In our case, it is ${\displaystyle \frac{x}{a+nx}}$ where $n\to \mathrm{\infty }$. So, it is zero, thus $\left|{\displaystyle \frac{x}{a+nx}}\right|<1$ so we can use the above expansion.