Approximating the convergent series $\sum_{n=1}^\infty\frac{(-1)^nn}{5^n}$

Question

I have the following series $\displaystyle \sum_{n=1}^\infty\frac{(-1)^nn}{5^n}$ which I have to estimate to within 0.0001.

But I don't know how to solve the inequality $\displaystyle \frac n{5^n}<0.0001$ to get $n$. Is this right? If so, how do I proceed?

Answer 1

$\frac n{5^n}<0.0001$ is the correct inequality to set up, and the smallest admissible value of $n$ will be the final one required to estimate the alternating sum to within the specified tolerance.

Noting that $0.0001=10^{-4}=\frac{5^{-4}}{16}$, we may rearrange the inequality: $$16n<5^{n-4}$$ and then test values of $n$ from 4 upwards until the inequality is satisfied. In this case, the smallest such $n$ is 7; the estimated sum then works out to be $-0.1389056$. (The exact infinite sum is $-\frac5{36}=-0.13\overline8$.)

Answer 2

Just added for your curiosity.

Making the problem more general, you are looking for the smallest $n$ such that $$\frac n{a^n} <\epsilon$$ The solution is given in terms of Lambert function $$n >-\frac{W_{-1}(-\epsilon \log (a))}{\log (a)}$$ The wikipedia page will show you the approximation $$W_{-1}(x)=L_1-L_2+\frac{L_2}{L_1}+\cdots$$ where $L_1=\log(-x)$ and $L_2=\log(-L_1)$.

Applied to your case ($a=5$ and $\epsilon=10^{-4}$), this would use $L_1\approx -8.73446$, $L_2\approx 2.16728$ making $n\approx 6.9278$ that is to say $n=7$ as Parcly Taxel already answered.

If you wanted lower tolerances, writing $\epsilon=10^{-k}$, you would get $$\left( \begin{array}{cc} k & n \\ 4 & 6.92508 \\ 5 & 8.48174 \\ 6 & 10.0157 \\ 7 & 11.5341 \\ 8 & 13.0411 \\ 9 & 14.5393 \\ 10 & 16.0307 \end{array} \right)$$ and you need to use the next integer value.