# Convergence Test Calculator + Online Solver With Free Steps

The Convergence Test Calculator is used to find out the convergence of a series. It works by applying a bunch of Tests on the series and finding out the result based on its reaction to those tests.

Calculating the sum of a Diverging Series can be a very difficult task, and so is the case for any series to identify its type. So, certain tests have to apply to the Function of the series to get the most appropriate answer.

## What Is a Convergence Test Calculator?

The Convergence Test Calculator is an online tool designed to find out whether a series is converging or diverging.

The Convergence Test is very special in this regard, as there is no singular test that can calculate the convergence of a series.

So, our calculator uses several different testing methods to get you the best result. We will take a deeper look at them as we move forward in this article.

## How to Use the Convergence Test Calculator?

To use the Convergence Test Calculator, enter the function of the series and the limit in their appropriate input boxes and press the button, and you have your Result. Now, to get the step-by-step guide to making sure you get the best results from your Calculator, look at the given steps:

### Step 1

We start by setting up the function in the appropriate format, as the variable is recommended to be n instead of any other. And then enter the function in the input box.

### Step 2

There are two more input boxes, and these are the ones for “to” and “from” limits. In these boxes, you are to enter the lower limit and the upper limit of your series.

### Step 3

Once, all of the above steps are completed, you may press the button labeled “Submit”. This will open up a new window where your solution will be provided.

### Step 4

Finally, if you wish to find out about more series’ convergence you can enter your new problems in the new window, and get your results.

## How Does the Convergence Test Calculator Work?

The Convergence Test Calculator works by testing a series to the limit of infinity and then concluding whether it’s a Convergent or Divergent series. This is important because a Convergent Series will converge to a certain value at some point at infinity, and the more we add the values into such a series the closer we get to that Certain Value.

Whilst, on the other hand, Divergent Series do not get a defined value as you add them, they instead diverge either into infinity or some random sets of values. Now, before we move forward to discuss how to find the Convergence of a series, let’s first discuss what a series is.

### Series

A Series in mathematics is referred to as a process rather than a quantity, and this Process involves adding a certain function to its values again and again. So, a series at its core is indeed a polynomial of some kind, with an Input variable that leads to an Output value.

If we apply a Summation function on top of this polynomial expression, we have a series limits of which are often approaching Infinity. So, a series could be expressed in the form:

$\sum_{n=1}^{\infty} f(n) = x$

Here, the f(n) describes the function with variable n and the output x could be anything from a defined value to Infinity.

### Convergent and Divergent Series

Now, we will investigate what makes a series Convergent or Divergent. A Convergent Series is one that when added up many times will result in a particular value. This value can be approached as a value of its own, so let our Convergent Series result in a number x after 10 iterations of the summation.

Then, after 10 more it will approach a value that would be not too far from x but a better approximation of the series’ result. An Important Fact to notice is that the result from more sums would be almost always Smaller than the one from lesser sums.

A Divergent Series on the other hand when added more times would usually result into a bigger value, which would keep increasing thus diverging that it would approach Infinity. Here, we have an example of each Convergent as well as Divergent Series:

$Convergent: \phantom {()} \sum_{n=1}^{\infty} \frac {1} {2^n} \approx 1$

$Divergent: \phantom {()} \sum_{n=1}^{\infty} 112 n \approx \infty$

### Convergence Tests

Now, to test the convergence of a series, we can use several techniques called Convergence Tests. But it must be noted that these tests only come into play when the Series’ Sum cannot be calculated. That occurs very commonly when dealing with values adding up to Infinity.

The first test we look at is called the Ratio Test.

1. #### Ratio Test

A Ratio Test is mathematically described as:

$\lim_{n\to\infty} \frac {a_{n + 1}} {a_n} = D$

Here, the subscripts describe the position of the number in the series, as an would be nth number, and a{n+1} would be $(n+1)^{th}$ number.

Where D is the most important value here, if it is less than 1, the series is Convergent, and if greater than 1 then otherwise. And if the value of D comes to be equal to 1, the test becomes incapable of answering.

But we won’t stop at just one test, and carry forward to another one called the Root Test.

1. #### Root Test

A Root Test can be mathematically described as:

$\lim_{n\to\infty} \sqrt[n]{a_n} = D$

And similar to the Ratio Test, an represents the value in the series at the point n. Where D is the determining factor if it is bigger than 1 the series is Divergent, and if smaller than 1 otherwise. And for equal to 1 the test becomes unreliable, and the answer becomes Inconclusive.

## Solved Examples

Now, let’s take a deeper look and get a better understanding of the concepts using some examples.

### Example 1

Consider the series expressed as:

$\sum_{n=0}^{\infty} \frac {n} {4^n}$

Find out if the series is convergent or not.

### Solution

We begin by first analyzing the series and checking if it’s possible to calculate its Sum. And as it is seen that the function contains the variable $n$ in both the Numerator and the Denominator. The only hint is that the denominator is in the form of an Exponential, but we may have to rely on a test for this.

So, we will first apply the Ratio Test on this series and see if we can get a viable result. First, we have to set up the values for the test, as the test is described as:

$\lim_{n\to\infty} \frac {a_{n + 1}} {a_n}$

$a_n = \frac {n} {4^n}, \phantom {()} a_{n+1} = \frac {n + 1} {4^{n + 1}}$

Now, we shall put this into the mathematical description of the test:

$\lim_{n\to\infty} \frac {a_{n + 1}} {a_n} = \lim_{n\to\infty} \frac {4^n \cdot (n + 1)} {n \cdot 4^{n + 1}} = \lim_{n\to\infty} \frac {n+1} {4 \cdot n}$

$\lim_{n\to\infty} \frac {n+1} {4 \cdot n} = \frac {1} {4} \cdot \lim_{n\to\infty} \bigg ( 1 + \frac {1}{n} \bigg ) = \frac {1} {4}$

As the answer is smaller than $1$, the series is convergent.

### Example 2

Consider the series given as:

$\sum_{n=0}^{\infty} \bigg( \frac {5 \cdot n + 1} {2 \cdot n + 5} \bigg) ^ {6 \cdot n + 2}$

Find whether the series is Convergent or Divergent.

### Solution

We start by looking at the series itself, and whether we can sum it up. And it’s very easily obvious that we can’t. The series is very complicated, so we must then rely on a test.

So, we shall use the Root Test for this, and see if we can get a viable result from it. We start off by setting up our problem according to the test requirements:

$\lim_{n\to\infty} \sqrt[n]{a_n}$

$a_n = \bigg( \frac {5 \cdot n + 1} {2 \cdot n + 5} \bigg) ^ {6 \cdot n + 2}$

Now, we will place the value of an into the mathematical description of the test:

$\lim_{n\to\infty} \sqrt[n]{a_n} = \lim_{n\to\infty} \sqrt[n]{ \bigg( \frac {5 \cdot n + 1} {2 \cdot n + 5} \bigg) ^ {6 \cdot n + 2}} = \lim_{n\to\infty} \bigg( \frac {5 \cdot n + 1} {2 \cdot n + 5} \bigg) ^ {\frac{6 \cdot n + 2} {n}} = \lim_{n\to\infty} \bigg( \frac { \frac{5 \cdot n + 1}{n}} {\frac{2 \cdot n + 5}{n}} \bigg) ^ {6 + \frac{2} {n}}$

$\lim_{n\to\infty} \bigg( \frac { \frac{5 \cdot n + 1}{n}} {\frac{2 \cdot n + 5}{n}} \bigg) ^ {6 + \frac{2} {n}} = \lim_{n\to\infty} \bigg( \frac { \frac{5 \cdot n + 1}{n}} {\frac{2 \cdot n + 5}{n}} \bigg) ^ {6} \cdot \lim_{n\to\infty} \bigg( \frac { \frac{5 \cdot n + 1}{n}} {\frac{2 \cdot n + 5}{n}} \bigg) ^ { \frac{2} {n}} = (\frac{5}{2})^6 = \frac{15625}{64}$

As the answer is greater than 1, so the series is divergent.