Series Calculator computes sum of a series over the given interval. It is capable of computing sums over finite, infinite and parameterized sequences. For the finite sums series calculator computes the answer quite literally, so if there is a necessity to obtain a short expression we recommend computing a parameterized sum. The Infinite Series Calculator an online tool, which shows Infinite Series for the given input.

### How to Use Series Calculator

Necessary condition for a numerical sequence convergence is that limit of common term of series is equal to zero, when the variable approaches infinity. However, this condition is not sufficient to determine the convergence of numerical series online. For series convergence determination a variety of sufficient criterions of convergence or divergence of a series have been found. If an input is given then it can easily show the result for the given number. A special place among numeric series is occupied by such in which the signs of the summands being strictly alternated, and absolute values of the numeric series monotonously subside.

I made a final set of calculator GCSE revision mats in case you need something for Saturday/Monday. As before, they're not for a specific exam board or series so they can be used every year. There's 4 different difficulty levels – one to suit every class. https://t.co/UNnZS728HU pic.twitter.com/XTgjAkWSN3

— Jo Morgan (@mathsjem) 6 июня 2019 г.

A classmate found it on the internet and through a series of cables, he was able to load it on my calculator. I wonder if the same sites are still up.

— Jonathan Feng (@jonfeng1) 18 июня 2019 г.

A calculator does not have a conscience therefore it can not believe in anything.

It is just a series of logic gates.

Therefore, I repeat, if it doesn’t believe in god, or christ, how can it calculate, in your opinion?— An Oak Tree (@oak_writer) 26 июня 2019 г.

What is sum of (5^n*x^(2n))/(6^(n+1)) from n=0 to infinity

This series is convergent not for all x.

Find exact sum:

Summation(1/(n^2) – 1/(n+3)^2), n=1 to infinity

Can you show the steps please?

Write out a few terms: (1/1^2-1/4^2)+(1/2^2-1/5^2)+(1/3^2-1/6^2)+(1/4^2-1/7^2) + …

As you can see most terms are cancelled out and you’re left only with 1/1^2 + 1/2^2 + 1/3^2 = 49/36

from 1 to infinity sum of 4/(k+2)^2 Gotta find if it converges. Im kinda lost

It converges.

Thanks for The Calculator , Nice Website

Find if this series converges for 1 to inf: (-1)^n tan^-1 ((n+sin(n!)) / (n^2+sin(n!)^2))?

This series is hard to find.

binomial(2n,n)x^n from 0 to infinity=(1-4x)^-1/2 now put x=-1/16

If x = -1/16, then binomial(2n,n)(-1/16)^n from 0 to infinity=(1-4(-1/16))^-1/2 = (1+1/4)^(-1/2)=(5/4)^(-1/2)=(4/5)^(1/2)=2/sqrt(5)

Your calculators were great but the advertisements you have are unbearable. Every time I click calculate, an ad pops up, and you have an ad on the bottom right corner that blocks the solutions. When I try to close it, another ad pops up. I literally get 7 ads that open up in new tabs for each time I use the calculator.

The way you had it before was great, you still had advertisements but it did not prevent me/us from calculating problems. Please fix this otherwise, it would not be useful to use.

I will look at it.

find the sum of (cos(2/n^2)-cos(2/(n+1)^2)) from 1 to infinity

cos(2) – 1

Very helpful website! Can you help me on the following:

sin(5n)/(1+sqrtn)

I know it converges, but to what?

What are the limits for n?

Determine the number of terms, n, given the geometric series -2 – 6 – 18 – 54 – … and Sn=−2186

The number of terms is 7.

Why the sum of ((1/3)^x)*x to infinite is equal to 3/4.

I know that the sum of geometric series (1/3)^x will converge to 3/2, with formula a/(1-(1/3)). But ((1/3)^x)*x is no longer geometric series. Therefore I don’t know the algothrim anymore. Can anyone show me how to get the answer?

A geometric series is given by the sum of a*n^x as x goes from 1 to infinity. Its sum is a*n/(1-n)

Differentiate both sides with respect to n: a*x*n^(x-1) as n goes from 1 to infinity equals a/(1-n)^2

Since n^(x-1) = n^x / n, then a*x*n^(x-1) = (a/n) x n^x and this sum equals a/(1-n)^2

Therefore, the sum of x n^x equals a/(1-n)^2 * (n/a) = n/(1-n)^2

In your case n=1/3: the sum is (1/3)/(1-1/3)^2=3/4

Great calculator! I was trying to find this sort all along to finish my homeworks

Thank you for your words!

can’t find the answer: sum from 1 to inf of (n!)/((n+1)!+n^(1/2))

The series diverges.

I have an infinite series that seems to have bested your website and was wondering whether it was solvable- I couldn’t find many series like it:

.1n(.9^(n-1))

The partial sum reaches its practical limit at 10 and the websites I check against agree that it converges, but I would like to know if there is a way to find the true limit. Thanks!

What does the dot sign mean?

That was a decimal point- .1n=0.1n, .9=0.9

The answer is 10. You want steps?

A geometric series is given by the sum of a*x^n as n goes from 1 to infinity. Its sum is a*x/(1-x)

Differentiate both sides: a*n*x^(n-1) as n goes from 1 to infinity equals a/(1-x)^2

You have 0.1n(0.9^(n-1)), so in your case a=0.1, x=0.9: the sum is 0.1/(1-0.9)^2=10