# How to find the error or remainder of a series

## The sum of the series is usually the sum of the first several terms, plus a very smaller error that you can estimate

Imagine that you need to find the sum of a series, but you don’t have a formula that you can use to do it. Instead, you have to manually add all of the series’ terms together, one at a time. Of course you could never do this, because the series has an infinite number of terms, and you’d be adding forever.

Hi! I'm krista.

But what if you knew that the sum of just the first five terms of the series was only ???.00001??? less than the sum of the entire series? If that were the case, maybe you could just use the first five terms, and say that it was a good enough estimate of the total sum, since it’s only ???.00001??? different and it saves you from manually adding infinitely more terms to the sum.

If you use the estimate, then you want to be able to report next to your answer that the value you found is only ???.00001??? off of the total sum. This ???.00001??? value is called the remainder, or error, of the series, and it tells you how close your estimate is to the real sum.

To find the remainder of the series, we’ll need to

1. Estimate the total sum by calculating a partial sum for the series.

2. Use the comparison test to say whether the series converges or diverges.

3. Use the integral test to solve for the remainder.

## Using the first six terms to estimate the remainder

Example

Use the first six terms to estimate the remainder of the series.

???\sum^{\infty}_{n=1}\frac{n}{2n^4+3}???

The first thing we need to do is to find the sum of the first six terms ???s_6??? of our original series ???a_n???.

The sum of the first six terms of the series ???a_n??? is

???s_6=\frac15+\frac{2}{35}+\frac{1}{54}+\frac{4}{515}+\frac{5}{1,253}+\frac{6}{2,595}???

???s_6=0.2000+0.0571+0.0185+0.0078+0.0040+0.0023???

???s_6=0.2897???

Since we’ve rounded our decimals, we’ll say

???s_6\approx0.2897???

Next, we need to use the comparison test to figure out whether ???a_n??? converges or diverges. We will need to create a similar but simpler comparison series ???b_n???. We can use the same numerator in ???b_n??? as the numerator from ???a_n???, since it’s already pretty simple. For the denominator, we can use ???n^4???, since it’s the element of the denominator that has the most impact on the series. The comparison series ???b_n??? will be

???b_n=\frac{n}{n^4}???

???b_n=\frac{1}{n^3}???

The comparison series ???b_n??? is a p-series where ???p=3???. The p-series test tells us that the series

will converge when ???p>1???

will diverge when ???p\le1???

Since ???p=3???, we know that ???b_n??? converges.

To use the comparison test to show that ???a_n??? also converges, we have to show that ???0\le{a_n}\le{b_n}???. We’ll find some of the first few values of the comparison series ???b_n??? and compare them to ???a_n???. Let’s use ???n=1,\ 2,\ 3???.

Looking at these three terms and their corresponding terms from ???a_n???, we can see that ???0\le{a_n}\le{b_n}???, which means that ???a_n??? converges.

You’ll Use the integral test to find the remainder of the series

Now that we know that the series converges, we’ll use the integral test to find the remainder of the series ???a_n??? after the first six terms, ???R_6???. We’ll call the remainder of the comparison series ???b_n??? after the first six terms, ???T_6???. Since we know that ???0\le{a_n}\le{b_n}???, and that ???a_n??? and ???b_n??? converge, we can say that ???R_6\le{T_6}???, which will be less than the total area under ???b_n???.

???R_6\le{T_6}\le\int^{\infty}_6b_n\ dx=\int^{\infty}_6f(x)\ dx???

???R_6\le{T_6}\le\int^{\infty}_6b_n\ dx=\int^{\infty}_6\frac{1}{x^3}\ dx???

???R_6\le{T_6}\le\int^{\infty}_6b_n\ dx=\int^{\infty}_6{x^{-3}}\ dx???

???R_6\le{\frac{x^{-2}}{-2}}\bigg|^{\infty}_6???

???R_6\le\lim_{a\to{\infty}}{\frac{x^{-2}}{-2}}\bigg|^a_6???

???R_6\le\lim_{a\to{\infty}}{-\frac{1}{2x^2}}\bigg|^a_6???

???R_6\le\lim_{a\to{\infty}}{-\frac{1}{2a^2}}-\left({-\frac{1}{2(6)^2}}\right)???

???R_6\le\lim_{a\to{\infty}}\frac{1}{2(6)^2}-\frac{1}{2a^2}???

???R_6\le\lim_{a\to{\infty}}\frac{1}{72}-\frac{1}{2a^2}???

???R_6\le\frac{1}{72}-\frac{1}{2\infty^2}???

???R_6\le\frac{1}{72}-0???

???R_6\le\frac{1}{72}???

???R_6\le0.0139???

The sixth partial sum of the series ???a_n??? is ???s_6\approx0.2897???, with error ???R_6\le0.0139???.