In calculus, convergence tests are nothing but the methods of testing for the convergence such as conditional convergence, interval convergence and absolute convergence or divergence of an infinite series. In this article, we are going to discuss about various convergence tests such as ratio test, root test, integral test, limit comparison test and Cauchy's tests.

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Let us take the series $\sum x_n$ and it's partial sum ${S_n}$. The series converges, that is $S_n$ converges. The total sum of the series is given by taking limit

$\lim_{n \to \infty} S_n = \sum_{n = 1}^{\infty} x_n$. The series converges if the sum is convergent. There are a series of tests which are used to find whether a series converges or not, like ratio test, root test, integral test, limit test etc.

### Geometric Series Convergence Test:

Any geometric series $\sum_{n = 0}^{\infty}ax^{n}$ converges to its limit whenever its ratio $x$ lies in between the interval $(-1,1)$.

### Convergence Tests for Infinite Series:

### Integral Test for Convergence Examples

Given below are some of the examples on Integral Test for Convergence.### Solved Example

**Question: **Determine the series $\sum_{n = 3}^{\infty}\frac{1}{n \ln^{2}n}$ is convergent or divergent.

** Solution: **

**First Comparison Test:** Let $\sum x_{n}$ and $\sum y_{n}$ be two positive term series such that $x_{n} \leq r y_{n} \forall n\geq m$ where, $r$ is a fixed positive number and $m$ is a fixed positive integer. Then,

### Comparison Test for Convergence Examples

### Solved Examples

**Question 1: **State whether the series is convergent or not by using any one of the above tests

$\sum_{n \geq 1} \frac{n^n}{n!}$

** Solution: **
**Question 2: **Check whether the series is convergent or not:

$\sum_{n = 0}^{\infty} (\frac{n}{2n + 1})^n$

** Solution: **
### Ratio Test for Convergence Examples:

If we have the series $\sum \frac{5^{n}}{n!}$, then the nth term and n + 1 th term of the series is given by $a_{n} = \frac{5^{n}}{n!}$ and $a_{n + 1} = \frac{5^{n + 1}}{(n + 1)!}$ respectively. Then, we can find the ratio of these two terms as follows:

$\frac{a_{n + 1}}{a_n} = \frac{\frac{5^{n + 1}}{(n + 1)!}} {\frac{5^{n}}{n!}}$

= $\frac{5^{n + 1}}{(n + 1)!} . \frac{n!}{5^{n}}$

= $ \frac{5}{n + 1} \rightarrow 0$

Here, the limit r = 0 < 1, then the given series is convergent.

### Ratio Test Radius of Convergence:

### Solved Example

**Question: **Determine the series $\sum_{n = 1}^{\infty}\frac{1}{n + 1}$ is convergent or divergent.

** Solution: **
If we have a series $\sum x_{n}$ and if the term $\sum \left | x_{n} \right |$ is convergent then the series $\sum x_{n}$ is called absolute convergent. If any series is absolute convergent, then that will also be convergent.

For example, if we have alternating Harmonic series $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$ which is convergent also. Then for absolute convergent we have to check

$\sum_{n=1}^{\infty}\left | \frac{(-1)^{n}}{n} \right |= \sum_{n=1}^{\infty}\frac{1}{n}$

With the help of integral test, it is clear that the above series is divergent and hence the given series is conditionally convergent not absolute convergent.

If the series $\sum u_{n}$ converges, then $\lim_{n \rightarrow \infty }u_{n} = 0$. But, the converse is not true. This is the necessary condition for convergence. So, we say that if $S_n = a_1 + a_2 + a_3 + ....... + a_n$ is the set of given $n^th$ partial sums series $\sum_{n = 1}^{\infty}u_{n}$.

If the given sequence $<S_n>$ converges to any limit $M$, then the sum of the series converges to that limit $M$ and if the sequence $<S_n>$ diverges, then the given series is divergent. In short, we can say that

If we have a series $\sum_{n = 2}^{\infty}\frac{1}{n^{2} - 1}$. To know about convergence or divergence of the series, let

$a_{n} = \frac{1}{n^{2} - 1}$ and $b_{n} = \frac{1}{n^{2}}$

Then, by limit test

$c = \lim_{n\rightarrow \infty }\frac{a_{n}}{b_{n}}$

= $\lim_{n\rightarrow \infty }\frac{\frac{1}{n^{2}-1}}{\frac{1}{n^{2}}}$

= $\lim_{n\rightarrow \infty }\frac{n^{2}}{n^{2}-1}$

= $\lim_{n\rightarrow \infty }\frac{1}{1-\frac{1}{n^{2}}}$

= $\frac{1}{1-0}$

= $1$

We have $c = 1$ and $\sum_{n = 2}^{\infty}\frac{1}{n^{2}}$ is convergence.

$\lim_{n \to \infty} S_n = \sum_{n = 1}^{\infty} x_n$. The series converges if the sum is convergent. There are a series of tests which are used to find whether a series converges or not, like ratio test, root test, integral test, limit test etc.

We compare the integral of the series to test whether it is a convergence or divergence series. Let us consider $f(1, \infty) \to R+$ as a positive function given that $f(n) = a_n$. If $\int_1^{\infty} f(x) dx = \lim_{t \to \infty} \int_1^t f(x) dx < \infty$, then the series is called as convergence series. When the integral is said to diverge, the series also diverges.

If f is a positive, continuous, decreasing function on the interval $[1, \infty]$ and assume $f(n) = x_n$. And, if the improper integral $\int_{1}^{\infty}f(x)dx$ is convergent, then the given series $\sum x_{n}$ is convergent or we say that $\sum x_{n}$ is convergent iff $\int_{1}^{\infty}f(x)dx$ is convergent.Given below are some of the examples on Integral Test for Convergence.

Since we know that the function $f(x) = \frac{1}{x \ln^{2}x}$ is positive, continuous, decreasing on the interval $[3, \infty]$.

Then, by integral test

$\int_{3}^{\infty} \frac{1}{x \ln^{2} x}dx = \lim_{s \rightarrow \infty }\int_{3}^{s} \frac{1}{x \ln^{2}x}dx$

= $\lim_{s \rightarrow \infty }\left [ -\frac{1}{\ln x} \right ]_{3}^{s}$

= $\lim_{s\rightarrow \infty }\left [ -\frac{1}{\ln s} + \frac{1}{\ln 3}\right ]$

= $\frac{1}{\ln 3}$

So, it is clear that the integral is convergent and hence the series $\sum_{n = 3}^{\infty}\frac{1}{n \ln^{2}n}$ is convergent.

Then, by integral test

$\int_{3}^{\infty} \frac{1}{x \ln^{2} x}dx = \lim_{s \rightarrow \infty }\int_{3}^{s} \frac{1}{x \ln^{2}x}dx$

= $\lim_{s \rightarrow \infty }\left [ -\frac{1}{\ln x} \right ]_{3}^{s}$

= $\lim_{s\rightarrow \infty }\left [ -\frac{1}{\ln s} + \frac{1}{\ln 3}\right ]$

= $\frac{1}{\ln 3}$

So, it is clear that the integral is convergent and hence the series $\sum_{n = 3}^{\infty}\frac{1}{n \ln^{2}n}$ is convergent.

- $\sum y_{n}$ converges $\Rightarrow \sum x_{n}$ converges.
- $\sum x_{n}$ diverges $\Rightarrow \sum y_{n}$ diverges.

**Second Comparison Test:** If $\sum x_{n}$ and $\sum y_{n}$ be two positive term series such that $\frac{x_{n}}{x_{n + 1}} \geq \frac{y_{n}}{y_{n + 1}} \forall n > m$. Then,

- $\sum y_{n}$ converges $\Rightarrow \sum x_{n}$ converges.
- $\sum x_{n}$ diverges $\Rightarrow \sum y_{n}$ diverges.

**Given below are some of the examples on comparison test for convergence.**

$\sum_{n \geq 1} \frac{n^n}{n!}$

We have a factorial notation. So, we use the ratio test.

$\frac{\frac{(n + 1)^{n + 1}}{(n + 1)!}}{\frac{n^n}{n!}} = (\frac{n + 1}{n})^n$

= $(1 + \frac{1}{n})^n$

Applying limits, we get

$\lim_{n \to \infty}(1 + \frac{1}{n})^n = e > 1$. As $e > 1$, this series diverges.

$\frac{\frac{(n + 1)^{n + 1}}{(n + 1)!}}{\frac{n^n}{n!}} = (\frac{n + 1}{n})^n$

= $(1 + \frac{1}{n})^n$

Applying limits, we get

$\lim_{n \to \infty}(1 + \frac{1}{n})^n = e > 1$. As $e > 1$, this series diverges.

$\sum_{n = 0}^{\infty} (\frac{n}{2n + 1})^n$

We use the root test

$\lim_{n \to \infty}(\frac{n}{2n + 1})^n = \lim_{n \to \infty} ((\frac{n}{2n + 1})^n)^{\frac{1}{n}}$

= $\lim_{n \to \infty} \frac{n}{2n + 1} = \frac{1}{2}$

The series converges.

$\lim_{n \to \infty}(\frac{n}{2n + 1})^n = \lim_{n \to \infty} ((\frac{n}{2n + 1})^n)^{\frac{1}{n}}$

= $\lim_{n \to \infty} \frac{n}{2n + 1} = \frac{1}{2}$

The series converges.

This is very important test to check the convergence of any power series. For this consider, for all n, a_{n }> 0 and there exists r such that

$\lim_{n\rightarrow \infty }\frac{a_{k + 1}}{a_{k}} = r$

If r < 1, then the series is said to converge.

If we have the series $\sum \frac{5^{n}}{n!}$, then the nth term and n + 1 th term of the series is given by $a_{n} = \frac{5^{n}}{n!}$ and $a_{n + 1} = \frac{5^{n + 1}}{(n + 1)!}$ respectively. Then, we can find the ratio of these two terms as follows:

$\frac{a_{n + 1}}{a_n} = \frac{\frac{5^{n + 1}}{(n + 1)!}} {\frac{5^{n}}{n!}}$

= $\frac{5^{n + 1}}{(n + 1)!} . \frac{n!}{5^{n}}$

= $ \frac{5}{n + 1} \rightarrow 0$

Here, the limit r = 0 < 1, then the given series is convergent.

Here, we can determine the interval and radius of convergence of the power series $\sum_{n=0}^{\infty}n! (x+1)^{n}$

For this we can use the ration test, hence we have $L = \lim_{n\rightarrow \infty }\left | \frac{(n+1)! (x

+1)^{n+1}}{n! (x+1)^{n}} \right |$

= $\lim_{n\rightarrow \infty }\left | (n+1)(x+1) \right |$

= $(x+1) \lim_{n\rightarrow \infty }(n+1)$

It is clear that there is x's in front of the limit and the limit if infinite, It gives L = $\infty$ > 1 only when x $\neq$ -1. So, the given power series is converge to its limit only if x = -1.

The power series $\sum_{n = 1}^{\infty}\frac{1}{n^{p}}$ is convergent if p > 1. Some times, we cannot apply the p-test directly, but we can use it after the integral test. In the given problem, we can not use p-test directly. Since the given function $f(x) = \frac{1}{x + 1}$ is positive, continuous, decreasing function on the interval $[1, \infty]$. Now, by the use of integral test,

$\int_{1}^{\infty } \frac{1}{x + 1}dx = \lim_{s \rightarrow \infty }\int_{1}^{s}\frac{1}{x + 1}dx$

= $\lim_{s \rightarrow \infty } \ln(x + 1)_{1}^{s}$

= $\lim_{s \rightarrow \infty }\left [ \ln(x + 1) - \ln 2 \right ]$

= $\infty$

Since the integral is divergent and hence the given series $\sum_{n = 1}^{\infty}\frac{1}{n + 1}$ is divergent.

Now, we have

$\sum_{n = 1}^{\infty}\frac{1}{n + 1} = \sum_{n = 2}^{\infty}\frac{1}{n}$

The term $\sum_{n = 2}^{\infty}\frac{1}{n}$ is divergent with $p = 1$. Hence, the given series is divergent.

$\int_{1}^{\infty } \frac{1}{x + 1}dx = \lim_{s \rightarrow \infty }\int_{1}^{s}\frac{1}{x + 1}dx$

= $\lim_{s \rightarrow \infty } \ln(x + 1)_{1}^{s}$

= $\lim_{s \rightarrow \infty }\left [ \ln(x + 1) - \ln 2 \right ]$

= $\infty$

Since the integral is divergent and hence the given series $\sum_{n = 1}^{\infty}\frac{1}{n + 1}$ is divergent.

Now, we have

$\sum_{n = 1}^{\infty}\frac{1}{n + 1} = \sum_{n = 2}^{\infty}\frac{1}{n}$

The term $\sum_{n = 2}^{\infty}\frac{1}{n}$ is divergent with $p = 1$. Hence, the given series is divergent.

For example, if we have alternating Harmonic series $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$ which is convergent also. Then for absolute convergent we have to check

$\sum_{n=1}^{\infty}\left | \frac{(-1)^{n}}{n} \right |= \sum_{n=1}^{\infty}\frac{1}{n}$

With the help of integral test, it is clear that the above series is divergent and hence the given series is conditionally convergent not absolute convergent.

If the series $\sum u_{n}$ converges, then $\lim_{n \rightarrow \infty }u_{n} = 0$. But, the converse is not true. This is the necessary condition for convergence. So, we say that if $S_n = a_1 + a_2 + a_3 + ....... + a_n$ is the set of given $n^th$ partial sums series $\sum_{n = 1}^{\infty}u_{n}$.

If the given sequence $<S_n>$ converges to any limit $M$, then the sum of the series converges to that limit $M$ and if the sequence $<S_n>$ diverges, then the given series is divergent. In short, we can say that

- If the term $\sum u_{n}$ converges, then $\lim_{n\rightarrow \infty }u_{n} = 0$.
- If $\lim_{n\rightarrow \infty }u_{n} \neq 0$ then, the term $\sum u_{n}$ will diverge.

If we have a series $\sum_{n = 2}^{\infty}\frac{1}{n^{2} - 1}$. To know about convergence or divergence of the series, let

$a_{n} = \frac{1}{n^{2} - 1}$ and $b_{n} = \frac{1}{n^{2}}$

Then, by limit test

$c = \lim_{n\rightarrow \infty }\frac{a_{n}}{b_{n}}$

= $\lim_{n\rightarrow \infty }\frac{\frac{1}{n^{2}-1}}{\frac{1}{n^{2}}}$

= $\lim_{n\rightarrow \infty }\frac{n^{2}}{n^{2}-1}$

= $\lim_{n\rightarrow \infty }\frac{1}{1-\frac{1}{n^{2}}}$

= $\frac{1}{1-0}$

= $1$

We have $c = 1$ and $\sum_{n = 2}^{\infty}\frac{1}{n^{2}}$ is convergence.

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