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Infinite Series Calculator + Online Solver With Free Steps

The Infinite Series Calculator finds the sum of an infinite series expressed as a function of the sequence index n up to infinity or over the range of values, n = [x, y].

The calculator supports several series: arithmetic, power, geometric, harmonic, alternating, etc. A mathematical series is the sum of all elements in a well-defined sequence of values.

The calculator also supports variables in the input other than n, which allows it to solve for power series that generally contain a variable. However, the summation takes priority over characters as k > n > characters in alphabetical order. Thus if the input has any number of variables and: 

  • Contains k and n, then the summation is over k. 
  • Does not contain k but contains n, then the summation is over n. 
  • Contains neither k nor n, then the summation is over the variable appearing first in alphabetical order. So if the variables p and x appear, the summation is over p.

For simplicity, we will only use n as the summation variable throughout.

What Is the Infinite Series Calculator?

The Infinite Series Calculator is an online tool that finds the sum $\mathbf{S}$ of a given infinite sequence $\mathbf{s}$ over the range $\mathbf{n = [x, \, y]}$ where $\mathbf{x, \, y \, \in \, \mathbb{Z}}$ and $\mathbf{n}$ is the sequence index. The infinite sequence must be provided as a function $\mathbf{a_n}$ of $\mathbf{n}$. 

One of x and y can also be $-\infty$ or $\infty$ respectively, in which case $s_n = s_\infty = s$. Note that if x = $\infty$, the calculator will hang, so make sure that $x \leq y$.

The calculator interface consists of three text boxes labeled:

  1. “Sum of”: The function an to sum over that expresses a series as a function of n.
  2. “From” and “to”: The range of the variable n over which the sum takes place. The initial value goes into the box labeled “From” and the final value into the one labeled “to.”

Given the above inputs, the calculator evaluates the following expression and displays the result:

\[ S_n = \sum_{n=x}^y a_n \]

If one of x to -$\infty$ or y to $\infty$, then this is an infinite sum:

\[ S_n = S_\infty = S \]

\[ \sum_{n \, = \, x}^\infty a_n \, \, \text{if} \, \, y \to \infty \]

\[ \sum_{n\,=\,-\infty}^y a_n \, \, \text{if} \, \, x \to -\infty \]

Notation Explained

For an infinite sequence:

\[ s = \left \{ 1, \, \frac{1}{2}, \, \frac{1}{4}, \, \frac{1}{8}, \, \ldots \right \} \]

The corresponding infinite series is:

\[ S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \]

And the required summation form is:

\[ S = \sum_{n \,= \,0}^\infty a_n = \sum_{n \, = \, 0}^\infty \frac{1}{2^n} \]

Here, $a_n = \frac{1}{2^n}$ represents the required form of the input series (as a function of sequence index n), and S depicts the summation output.

How To Use the Infinite Series Calculator

You can use the Infinite Series Calculator by using the following guidelines. Suppose we want to find the infinite sum of the function:

\[ f(n) = a_n = \frac{3^n+1}{4^n} \]

That depicts some series over a range of n.

Step 1

Convert the sequence into a series and then the series into the summation form. If you already have the summation form, skip this step. In our case, we skip this step because we already have the summation form.

Step 2

Enter the series in the “Sum of” text box. For our example, we type “(3^n+1)/4^n” without commas.

Step 3

Enter the initial value for the summation range in the “From” text box. In our case, we type “0” without commas.

Step 4

Enter the final value for the summation range in the “to” text box. We type “infinity” without commas for our example, which the calculator interprets as $\infty$.

Step 5

Press the Submit button to get the results.

Results

Depending on the input, the results will be different. For our example, we get:

\[ \sum_{n \, = \, 0}^\infty \frac{3^n+1}{4^n} = \frac{16}{3} \, \approx \, 5.3333 \]

Infinite Range Sum

If the range of $n = [x, \, y]$ involves $x \, \, \text{or} \, \, y = \infty \, \, \text{or} \, \, -\infty$, the calculator perceives the input as a sum to infinity. This was the case with our mock example.

If the series diverges, the calculator will either show “the sum does not converge” or “diverges to $\infty$.” Otherwise, it displays the value on which the series converges. Our example input falls in this category.

Non-geometric Divergent Series

If you enter the function for an arithmetic series “1n” into the text box and evaluate it from 0 to infinity, the result will have an additional option “Show tests.” Clicking on that will present a list of five tests with their results that showed the series to be divergent. 

These tests are applied only when a direct method or formula such as the infinite sum of geometric series is not applicable. So for the input “2^n” (a function representing a geometric series over n), the calculator does not use these tests.

Finite Range Sum

If the range is well-defined and finite (e.g., $\sum_{n \, = \, 0}^5$), the calculator directly calculates the sum and displays it.

If the input sequence is one with a known closed form solution (arithmetic, geometric, etc.), the calculator uses it for a quick calculation.

How Does the Infinite Series Calculator Work?

The Infinite series calculator works by using the concept of sequences and series. Let’s have an insight into all the concepts involved in order to have a better understanding of the working of this calculator.

Sequences and Series

A sequence is a group of values where each element of the group is related to the next one in the same way. Extending such a group to infinity makes it an infinite sequence. For example:

\[ s_n = 1, \, \frac{1}{2}, \, \frac{1}{4}, \, \frac{1}{8}, \, \ldots \]

In the sequence above, if you pick the element si, you can determine $s_{i+1}$ by simply multiplying $s_i$ by $\frac{1}{2}$. Thus, each element in the sequence is half of the previous element.

\[ s_{i+1} = s_i \times \frac{1}{2} \]

We can find the value of any element in this sequence if we have one of the elements and its position/index. If we now sum all the elements of the sequence together, we get an infinite series:

\[ S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \]

Note that this particular series is known as the geometric series, where each consecutive term is related by a common ratio:

\[ r = \frac{a_{n+1}}{a_n} \]

Convergence and Divergence of Series

An infinite series can either converge (approach a definite, finite value) or diverge (approach an indefinite, infinite value). It may seem like an impossible problem, but we can perform several tests to determine whether a given series is convergent or divergent. The calculator uses the following:

  1. p-series Test
  2. Root Test
  3. Ratio Test
  4. Integral Test
  5. Limit/Divergence Test

In some cases, some of the tests might be inconclusive. Further, some tests indicate convergence but do not provide the convergence value. 

There are also techniques specific to types of series, such as for a geometric series with common ratio r:

\[ S_n = a + ar + ar^2 + \ldots + ar^{n-1} \]

We have the formula for the sum up to of n terms of the series:

\[ S_n = a \left ( \frac{1-r^{n+1}}{1-r} \right ) \, \, \text{where} \, \, r \neq 1 \]

If r > 1, the infinite geometric series is divergent since the numerator $a(1-r^{n+1}) \to \infty$ as $n \to \infty$. However, if r < 1, then the series is convergent and the formula simplifies to:

\[ S = \frac{a}{1-r} \, \, \text{if} \, \, r < 1 \]

Solved Examples

Example 1

Show that the harmonic series is divergent.

\[ H = \left\{ a + \frac{1}{a+d} + \frac{1}{a+2d} + \frac{1}{a+3d} + \ldots \right\} \]

Solution

The summation form of the series at a, d=1 is:

\[ H = \sum_{n \, = \, 1}^\infty \frac{1}{n} \]

The limit test is inconclusive as $\lim_{n \to \infty} \frac{1}{n} = 0$ and it is only valid for limiting values greater than 0.

The p-test states that for a sum of the form $\sum_{n \, = \, 1}^\infty \frac{1}{n^k}$, the series is divergent if $k \leq 1$ and convergent if k > 1. Here, the former is true so the series is divergent.

The integral test further validates the p-series result:

\[ \int_1^\infty \frac{1}{n} \cdot dn = \left. \ln n \right \rvert_1^\infty = \ln \infty \]

So the series is divergent.

Example 2

Evaluate:

\[ S = \sum_{n \, = \, 0}^\infty \frac{3^n+1}{4^n} \]

Solution

Let $a_n = \frac{3^n+1}{4^n}$. Breaking it into two fractions:

\[ a_n = \frac{3^n}{4^n} + \frac{1}{4^n} \]

Then our sum is essentially the sum of two geometric series:

\[ S = \underbrace{ \sum_{n \, = \, 0}^\infty \left ( \frac{3}{4} \right)^n }_\text{1$^\text{st}$ geometric series $G$} + \underbrace{ \sum_{n \, = \, 0}^\infty \left ( \frac{1}{4} \right)^n}_\text{2$^\text{nd}$ geometric series $G’$} \]

Where $r = \frac{3}{4} = 0.75 < 1$ for $G$ and $r’ = \frac{1}{4} = 0.25 < 1$ for G’, so both are convergent. Knowing that:

\[ a = \left. \left( \frac{3}{4} \right)^n \right \rvert_{n \, = \, 0} = 1 \]

\[ a’ = \left. \left( \frac{1}{4} \right)^n \right \rvert_{n \, = \, 0} = 1 \]

Using the infinite geometric sum formula:

\[ G = \frac{a}{1-r} = \frac{1}{0.25} = 4 \]

\[ G’ = \frac{a’}{1-r’} = \frac{1}{0.75} = \frac{4}{3} \]

\[ S = G + G’ = 4 + \frac{4}{3} = \frac{16}{3} \]

So the series is convergent.

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