Series evaluated to $m$ terms, approximating the error Given a series $\displaystyle\sum_{n=0}^\infty a_n$, how can we bound the error (which I shall denote with $R_n$) when we evaluate it to $m$ terms? 
$$\sum_{n=0}^\infty a_n \approx \sum_{n=0}^m a_n$$
$$R_n = \sum_{n = m+1}^\infty a_n$$ $$So$$
$$\sum_{n=0}^\infty a_n = \sum_{n=0}^m a_n + R_n$$
Clearly, we need to find some bounds for $R_n$ that we can evaluate, but how can they be found?

Example series:


*

*$\displaystyle\sum_{n=0}^\infty \frac1{n!}$

*$\displaystyle\sum_{n=0}^\infty \frac{2n+2}{(2n+1)!}$

*$\displaystyle\sum_{n=1}^\infty \frac1{n^2}$

*$\displaystyle\sum_{n=1}^\infty (-1)^n \frac1n$

*$\displaystyle\sum_{n=0}^\infty \frac{3+2n}{2^n}$





*

*For $\sum_{n=1}^\infty \frac1{n^2}$, we can use the integral test: $\displaystyle \int_{n+1}^\infty \frac1{x^2}dx \le R_n \le \int_{n}^\infty \frac1{x^2}dx$ so an upper bound on $R_n$ is $\displaystyle\lim_{t\to\infty}\left( \frac1n - \frac1t  \right) = \frac1n$

*For $\sum_{n=1}^\infty (-1)^n \frac1n$, we can use the property of alternating series that says $R_n \le \left|b_{n+1}\right|$

*For $\displaystyle\sum_{n=m}^\infty \frac{3+2n}{2^n}$, we can split it into $\displaystyle\sum_{n=m}^\infty\left(\frac3{2^n}+\frac{2n}{2^n}\right) = 2^{1-m} +\sum_{n=m}^\infty \frac{n}{2^{n-1}} < 2^{1-m} + \sum_{n=m}^\infty \frac {1.5^{n-1}}{2^{n-1}} = 2^{1-m} + \frac{3^{m-1}}{4^{m-2}}$ so $R_m \le 2^{1-m} + \frac{3^{m-1}}{4^{m-1}}$ given $m>5$.
 A: Typically one approaches these problems with some sort of comparison to a known series.  For example, 
$$\sum_{n=m}^\infty \frac{1}{n!}=\frac{1}{m!}\left(\frac{1}{m+1}+\frac{1}{(m+1)(m+2)}+\frac{1}{(m+1)(m+2)(m+3)}+\cdots\right)\le \frac{1}{m!}\left(\frac{1}{m+1}+\frac{1}{(m+1)^2}+\frac{1}{(m+1)^3}+\cdots\right)$$$$=\frac{1}{m!}\frac{\frac1{m+1}}{1-\frac1{m+1}}=\frac{1}{m!}\frac{m+1}{m}\le \frac{2}{m!}$$
 The known series in this case was a geometric series.
A: With alternating series of decreasing terms the first term left out is a bound for the error.
If your $a_n$ is a "nice" function $a(x)$ that is monotonically decreasing after the relevant $m$, you can use:
$$
\sum_{k \ge m} a(k) 
  = \int_m^\infty \lceil a(x) \rceil d x 
  \ge \int_m^\infty a(x) d x
  \ge \int_m^\infty \lfloor a(x) \rfloor d x
  = \sum_{k \ge m} a(k + 1)
  = \sum_{k \ge m + 1} a(k)
$$
Now estimate the integral somehow.
A: To find an estimate for the Error, we first try to use the integral test: $$\int_{n+1}^\infty a_n dn \le R_n \le \int_n^\infty a_n dn$$ or the alternating series test: $$R_n \le |b_{n+1}|$$ If we cannot do this, we then find another series to compare it to, one that we can evaluate, which is more than the original series: $$R_n = \sum_{m=n+1}^\infty a_n \le \sum_{m = n+1}^\infty b_n$$ and then evaluate the second series.
