Series evaluated to $m$ terms, approximating the error

Question

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?

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Example series:

• $\displaystyle\sum_{n=0}^\infty \frac1{n!}$
• $\displaystyle\sum_{n=0}^\infty \frac{2n+2}{(2n+1)!}$
• $\displaystyle\sum_{n=1}^\infty \frac1{n^2}$
• $\displaystyle\sum_{n=1}^\infty (-1)^n \frac1n$
• $\displaystyle\sum_{n=0}^\infty \frac{3+2n}{2^n}$

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• 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 $m>5$.

Answer 1

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.

Answer 2

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.

Answer 3

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.