For which positive integers k is the following series convergent?


This question aims to find the value of the positive integer $k$, for which the given series is convergent.

A series in mathematics is a representation of the procedure of adding infinite quantities sequentially to a given starting quantity. The series analysis is an important part of calculus and its generalization such as mathematical analysis. A convergent series is one in which the partial sums approach a particular number usually known as a limit. A divergent series is one in which the partial sums do not tend to a limit. Divergent series usually tend to positive or negative infinity and do not tend to a particular number.

The Ratio Test aids in determining whether a series converges or diverges. Consider the series $\sum a_n$. The Ratio Test examines $\lim\limits_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$ to determine the long term behavior of the series. As $n$ approaches infinity, this ratio compares the value of $a_{n+1}$ to the previous term $a_n$ to determine the amount of decrease in terms. If this limit is more than one, then $\left|\dfrac{a_{n+1}}{a_n}\right|$ will show that the series is not decreasing for all values of $n$ after a particular point. In this case, the series is said to be divergent. However, if this limit is smaller than one, absolute convergence can be observed in the series.

Expert Answer

Since the series is convergent, so by the Ratio Test:


$=\dfrac{[(n+1)!]^2}{[k(n+1)]!}\times \dfrac{(kn)!}{(n!)^2}$

$=\dfrac{[(n+1)\cdot n!]^2}{(kn+k)!}\times \dfrac{(kn)!}{(n!)^2}$

$=\dfrac{(n+1)^2\cdot (n!)^2}{(kn+k)\cdots (kn+2)\cdot (kn+1)(kn)!}\times \dfrac{(kn)!}{(n!)^2}$

$=\dfrac{(n+1)^2}{(kn+k)\cdots (kn+2)\cdot (kn+1)}$

Now, for $k=1$:


And so,  $\lim\limits_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|=\lim\limits_{n\to\infty}(n+1)=\infty$

Hence, the series diverges for $k=1$.

For $k=2$ we have:


And,  $\lim\limits_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|=\lim\limits_{n\to\infty}\dfrac{n^2+2n+1}{4n^2+6n+2}=\dfrac{1}{4}<1$

Hence, the series converges for $k=2$. We will have a function where the degree of the numerator will be smaller than the degree of the denominator for $k>2$. So, the limit becomes $0$ for $n$ approaching to $\infty$. Finally, it can be concluded that the given series converges for all $k\geq 2$.

Example 1

Determine whether the series  $\sum\limits_{n=1}^{\infty}\dfrac{(-15)^n}{3^{n+2}n}$ converges or diverges.


Let $a_n=\dfrac{(-15)^n}{3^{n+2}n}$

So, $a_{n+1}=\dfrac{(-15)^{n+1}}{3^{n+3}(n+1)}$

Suppose that  $L=\lim\limits_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$

$L=\lim\limits_{n\to\infty}\left|\dfrac{(-15)^{n+1}}{3^{n+3}(n+1)}\cdot \dfrac{3^{n+2}n}{(-15)^n}\right|$










So by Ratio Test, the given series is divergent.

Example 2

Test the series  $\sum\limits_{n=1}^{\infty}\dfrac{n!}{2^n}$, for convergence or divergence.


Let $a_n=\dfrac{n!}{2^n}$

So, $a_{n+1}=\dfrac{(n+1)!}{2^{n+1}}$

Let  $L=\lim\limits_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$

$L=\lim\limits_{n\to\infty}\left|\dfrac{(n+1)!}{2^{n+1}}\cdot \dfrac{2^n}{n!}\right|$

$L=\lim\limits_{n\to\infty}\left|\dfrac{(n+1)n!}{2^n\cdot 2^1}\cdot \dfrac{2^n}{n!}\right|$



Since the limit is equal to infinity, therefore, the given series is divergent by Ratio Test.

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