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For a convergent geometric series or telescoping series, we can find the exact error made when approximating the infinite series using the sequence of partial sums.

We’ve seen that if we have a convergent series , we can approximate its value by summing only finitely many of its terms. For instance if we consider $\sum _{k=1}^{\infty } a_k$, we can split the series up.

We thus have two important sequences associated with any convergent series $\sum _{k=n_0}^{\infty } a_k$.

• $\{s_n\}_{n=n_0}$ : The term $s_n$ is found by adding all terms up to and including $a_n$ and tells us the approximate value of the infinite series $\sum _{k=n_0}^{\infty } a_k$.
• $\{r_n\}_{n=1}$ : The term $r_n$ is the error made by approximating the infinite series $\sum _{k=n_0}^{\infty } a_k$ by $s_n$ (that is, by approximating $\sum _{k=n_0}^{\infty } a_k$ by $\sum _{k=n_0}^n a_k$).

When we have an explicit formula for $s_n$, we can determine an exact formula for the error term $r_n$; we simply use the relationship $\sum _{k=n_0}^{\infty } a_k = s_n +r_n$. We can compute $\sum _{k=n_0}^{\infty } a_k$ by taking $\lim _{n \to \infty } s_n$ and subtract $s_n$ form both sides to obtain the formula for $r_n$. We have seen that there are two special types of series when we can find such a description for $\{s_n\}_{n=1}$, which we explore below.

### Remainders for geometric series

Recall that a series $\sum _{k=n_0}^{\infty } a_k$ is geometric if for each $n\geq n_0$, there are constants $a$ and $r$ for which $a_n = ar^n$. We also know precisely when such series converge.

En route to establishing this result, we determined that when $n_0=0$,

We can use this to find a formula for $r_n$ when $|r|<1$.

Armed with an explicit formula for both $s_n$ and $r_n$, we can arrange the first several terms in each sequence in a table.

In most applications, we will only want to determine the value of a convergent series to a specified degree of accuracy. While we can compute the exact value of the series in the last example, this is not always the case. To gain some insight, we will use the previous exercise to estimate the series $\sum _{k=0}^{\infty } 2\left (\frac {1}{3}\right )^k$ to within $.001$ of its exact value. To do this, we will add finitely many terms and verify how good this approximation is.

### Remainders for telescoping series

The other type of series for which we could find an explicit formula for $s_n$ are telescoping series.

### Conclusion

When we have a convergent geometric or telescoping series, we can find an explicit formula for $s_n$, which allows us to find the exact value of the series as well as an explicit formula for the terms in the sequence of remainders. We have then seen how the terms in the sequence of remainders give the error made by approximating the infinite series by a finite sum.

One very important point to remember is that the calculations in this section are possible because we consider convergent series; remember it is not possible to define a sequence of remainders for a divergent series!

Select all of the series below for which it is possible to define a sequence of remainders.
$\sum _{k=2}^{\infty } \left (\frac {2}{9}\right )^k$ $\sum _{k=1}^{\infty } \left (\frac {4}{3}\right )^k$ $\sum _{k=1}^{\infty } \frac {2k}{4k+1}$