# Errors are decreasing in series $\sum_{n=1}^\infty(-1)^n/n^4$?

Let $$v=\sum_{n=1}^\infty(-1)^n/n^4$$ ($$v$$ for "value"), let $$S=(\sum_{n=1}^m(-1)^n/n^4)_{m\in\mathbb Z_{\ge1}}$$ be the partial sums, and let $$e=(|S_n-v|)_{n\in\mathbb Z_{\ge1}}$$ be the errors. Also note $$v=-\eta(4)=-7\pi/720$$ (Dirichlet eta).$$^1$$ My computations$$^2$$ of the first few errors suggest that the errors are strictly decreasing: $$e_1>e_2>e_3>\dots.$$ How can we prove this?

(I came up with this conjecture while solving a homework exercise for my undergraduate numerical analysis class. My conjecture is NOT part of the homework; I am asking out of curiosity. Also note $$v=\operatorname{Li}_4(-1)$$ (polylogarithm), if that helps.)

$$^1$$The proof that $$\eta(4)=7\pi^4/720,$$ or that $$v=-7\pi^4/720,$$ can be reduced to the proof that $$\zeta(4)=\pi^4/90.$$ For instance, let $$u=\sum_{n=1}^\infty n^{-4}.$$ Then $$v+u=\sum_{n=1}^\infty(-1)^nn^{-4}+\sum_{n=1}^\infty n^{-4}=\sum_{n=1}^\infty((-1)^n+1)n^{-4}=2\sum_{n=1}^\infty(2n)^{-4}=u/8.$$ Since $$u=\pi^4/90$$, we thus have $$v=-7u/8=-7\pi^4/720.$$ (Proof taken (almost) verbatim from ProofWiki.)

$$^2$$I used Matlab to compute the following errors. For example, $$\operatorname{round}_4(e_1)=5.30\!\cdot\!10^{-2},\operatorname{round}_5(e_2)=9.53\!\cdot\!10^{-3},\operatorname{round}_5(e_3)=2.81\!\cdot\!10^{-3},$$ where for any real number $$x$$ and any integer $$n\ge0,$$ $$\operatorname{round}_n(x)$$ is $$x$$ rounded to $$n$$ decimals.

First 150 errors: $$5.30\!\cdot\!10^{-2},9.53\!\cdot\!10^{-3},2.81\!\cdot\!10^{-3},1.09\!\cdot\!10^{-3},5.07\!\cdot\!10^{-4},2.65\!\cdot\!10^{-4},1.51\!\cdot\!10^{-4},9.27\!\cdot\!10^{-5},5.98\!\cdot\!10^{-5},4.02\!\cdot\!10^{-5},2.81\!\cdot\!10^{-5},2.02\!\cdot\!10^{-5},1.49\!\cdot\!10^{-5},1.12\!\cdot\!10^{-5},8.57\!\cdot\!10^{-6},6.68\!\cdot\!10^{-6},5.29\!\cdot\!10^{-6},4.24\!\cdot\!10^{-6},3.44\!\cdot\!10^{-6},2.81\!\cdot\!10^{-6},2.33\!\cdot\!10^{-6},1.94\!\cdot\!10^{-6},1.63\!\cdot\!10^{-6},1.38\!\cdot\!10^{-6},1.18\!\cdot\!10^{-6},1.01\!\cdot\!10^{-6},8.71\!\cdot\!10^{-7},7.56\!\cdot\!10^{-7},6.58\!\cdot\!10^{-7},5.76\!\cdot\!10^{-7},5.07\!\cdot\!10^{-7},4.47\!\cdot\!10^{-7},3.96\!\cdot\!10^{-7},3.52\!\cdot\!10^{-7},3.14\!\cdot\!10^{-7},2.81\!\cdot\!10^{-7},2.52\!\cdot\!10^{-7},2.27\!\cdot\!10^{-7},2.05\!\cdot\!10^{-7},1.86\!\cdot\!10^{-7},1.68\!\cdot\!10^{-7},1.53\!\cdot\!10^{-7},1.39\!\cdot\!10^{-7},1.27\!\cdot\!10^{-7},1.17\!\cdot\!10^{-7},1.07\!\cdot\!10^{-7},9.81\!\cdot\!10^{-8},9.03\!\cdot\!10^{-8},8.32\!\cdot\!10^{-8},7.68\!\cdot\!10^{-8},7.10\!\cdot\!10^{-8},6.58\!\cdot\!10^{-8},6.10\!\cdot\!10^{-8},5.66\!\cdot\!10^{-8},5.27\!\cdot\!10^{-8},4.90\!\cdot\!10^{-8},4.57\!\cdot\!10^{-8},4.27\!\cdot\!10^{-8},3.99\!\cdot\!10^{-8},3.73\!\cdot\!10^{-8},3.49\!\cdot\!10^{-8},3.27\!\cdot\!10^{-8},3.07\!\cdot\!10^{-8},2.89\!\cdot\!10^{-8},2.71\!\cdot\!10^{-8},2.56\!\cdot\!10^{-8},2.41\!\cdot\!10^{-8},2.27\!\cdot\!10^{-8},2.14\!\cdot\!10^{-8},2.02\!\cdot\!10^{-8},1.91\!\cdot\!10^{-8},1.81\!\cdot\!10^{-8},1.71\!\cdot\!10^{-8},1.62\!\cdot\!10^{-8},1.54\!\cdot\!10^{-8},1.46\!\cdot\!10^{-8},1.39\!\cdot\!10^{-8},1.32\!\cdot\!10^{-8},1.25\!\cdot\!10^{-8},1.19\!\cdot\!10^{-8},1.13\!\cdot\!10^{-8},1.08\!\cdot\!10^{-8},1.03\!\cdot\!10^{-8},9.80\!\cdot\!10^{-9},9.35\!\cdot\!10^{-9},8.93\!\cdot\!10^{-9},8.53\!\cdot\!10^{-9},8.15\!\cdot\!10^{-9},7.79\!\cdot\!10^{-9},7.45\!\cdot\!10^{-9},7.13\!\cdot\!10^{-9},6.83\!\cdot\!10^{-9},6.54\!\cdot\!10^{-9},6.27\!\cdot\!10^{-9},6.01\!\cdot\!10^{-9},5.76\!\cdot\!10^{-9},5.53\!\cdot\!10^{-9},5.31\!\cdot\!10^{-9},5.10\!\cdot\!10^{-9},4.90\!\cdot\!10^{-9},4.71\!\cdot\!10^{-9},4.53\!\cdot\!10^{-9},4.36\!\cdot\!10^{-9},4.19\!\cdot\!10^{-9},4.04\!\cdot\!10^{-9},3.89\!\cdot\!10^{-9},3.74\!\cdot\!10^{-9},3.61\!\cdot\!10^{-9},3.48\!\cdot\!10^{-9},3.35\!\cdot\!10^{-9},3.23\!\cdot\!10^{-9},3.12\!\cdot\!10^{-9},3.01\!\cdot\!10^{-9},2.91\!\cdot\!10^{-9},2.81\!\cdot\!10^{-9},2.71\!\cdot\!10^{-9},2.62\!\cdot\!10^{-9},2.54\!\cdot\!10^{-9},2.45\!\cdot\!10^{-9},2.37\!\cdot\!10^{-9},2.29\!\cdot\!10^{-9},2.22\!\cdot\!10^{-9},2.15\!\cdot\!10^{-9},2.08\!\cdot\!10^{-9},2.02\!\cdot\!10^{-9},1.95\!\cdot\!10^{-9},1.89\!\cdot\!10^{-9},1.83\!\cdot\!10^{-9},1.78\!\cdot\!10^{-9},1.72\!\cdot\!10^{-9},1.67\!\cdot\!10^{-9},1.62\!\cdot\!10^{-9},1.57\!\cdot\!10^{-9},1.53\!\cdot\!10^{-9},1.48\!\cdot\!10^{-9},1.44\!\cdot\!10^{-9},1.40\!\cdot\!10^{-9},1.36\!\cdot\!10^{-9},1.32\!\cdot\!10^{-9},1.28\!\cdot\!10^{-9},1.25\!\cdot\!10^{-9},1.21\!\cdot\!10^{-9},1.18\!\cdot\!10^{-9},1.15\!\cdot\!10^{-9},1.12\!\cdot\!10^{-9},1.09\!\cdot\!10^{-9},1.06\!\cdot\!10^{-9},1.03\!\cdot\!10^{-9},1.00\!\cdot\!10^{-9},9.74\!\cdot\!10^{-10}$$

• The is a well know result about alternating series, such as the one you are studying, that says that $|s-\sum^n_{k=1}(-1)^{k+1}a_k|<a_{n+1}$. Here $a_n\searrow0$. One can also fins the exact error and see is deacreasing. (See Apostol's analysis book, second edition pp 188-189.) Sep 20, 2020 at 17:03

In terms of the Hurwitz zeta function your $$m$$th error is \begin{align*} \left| {\sum\limits_{n = m + 1}^\infty {\frac{{( - 1)^n }}{{n^4 }}} } \right| & = \left| { \sum\limits_{n = 0}^\infty {\frac{{( - 1)^n }}{{(n + m + 1)^4 }}} } \right| \\ &= \frac{1}{{16}}\left( {\zeta \left( {4,\tfrac{{m + 1}}{2}} \right) - \zeta \left( {4,\tfrac{m}{2} + 1} \right)} \right) \\ & = \frac{1}{{96}}\int_0^{ + \infty } {t^3 e^{ - (m + 1)t/2} \frac{{1 - e^{ - t/2} }}{{1 - e^{ - t} }}dt} , \end{align*} showing that the error is strictly decreasing with respect to $$m$$.

$$v$$ is an alternating series. It is a general result regarding alternating series, that the error of the partial sum is bounded by the next general term. $$\frac{1}{(n+1)^4}$$ in this specific case.

See wikipedia article for more details and a proof of this result.

Also for an alternating series $$S= \sum_{n=0}^\infty \frac{(-1)^n}{a_n}$$ the partial sums $$S_k= \sum_{n=0}^k \frac{(-1)^n}{a_n}$$ are such that $$\{S_{2k}\}$$ is an increasing sequence. Hence the error $$\{e_{2k}\}$$ is a decreasing sequence. Similarly $$\{S_{2k+1}\}$$ is a decreasing sequence and $$\{e_{2k+1}\}$$ is also a decreasing sequence.

In our case we have

$$\begin{cases} e_{2k+1} &= S_{2k+1} - S = \sum_{n=2k+2}^\infty \frac{(-1)^n}{a_n}\\ e_{2k} &= S -S_{2k} = -\sum_{n=2k+1}^\infty \frac{(-1)^n}{a_n} \end{cases}$$

Therefore \begin{aligned}e_{2k}-e_{2k+1} &= 2S - S_{2k} - S_{2k+1}\\ &= \frac{1}{(2k+1)^4} - 2 \sum_{n=2k+2}^\infty \frac{(-1)^n}{n^4}\\ &= \frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right) \end{aligned}

Remains to prove that this sequence is non negative.

Which is done here by Martin R.

• Also that the errors are decreasing. Sep 20, 2020 at 15:50
• @mathcounterexamples.net How does this answer my question? Do we have $n^{-4}\le e_n$ or something? Sep 20, 2020 at 15:54
• You have $e_n < 1/(n+1)^4$. Sep 20, 2020 at 15:54
• How do you get from $e_n<1/(n+1)^4$ to $\dots<e_{n+2}<e_{n+1}<e_n<\dots<e_1$? Sep 20, 2020 at 15:56
• @mathcounterexamples.net Be careful, working with alternating series abstractly will not suffice here (on its own). For instance, you might ask: Are the errors (strictly) decreasing in the series $1-1+2^{-1}-2^{-1}+3^{-1}-3^{-1}+\dots$? The answer is "no": The even indices $n$ have the property $e_n=0.$ Sep 20, 2020 at 16:13

Using the approach from Proving that $\frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$ is non negrative we have the following general result:

If $$(a_n)$$ is a decreasing and convex sequence with $$\lim_{n\to \infty} a_n = 0$$ then the absolute values of series remainders (error terms) of the alternating series $$\sum_n (-1)^n a_n$$ are decreasing .

If the sequence is strictly convex then the absolute values of the series remainders are strictly decreasing.

A sequence is convex if $$a_{n+1} \le (a_{n} + a_{n+2})/2$$ for all $$n$$. This is satisfied in our case because $$f(x) = 1/x^4$$ is a convex function.

Proof of the above statement: Let $$e_n = \left | \sum_{j=n}^\infty (-1)^j a_j \right|= a_n - a_{n+1} + a_{n+2} - a_{n+3} + a_{n+4} - \ldots$$ denote the absolute value of the $$n^\text{th}$$ series remainder. Then $$e_{n+1} - e_{n} = -a_n +2 a_{n+1} -2 a_{n+2} +2 a_{n+3} - 2 a_{n+4} + \ldots\\ = (-a_n +2 a_{n+1} - a_{n+2}) + (- a_{n+2} +2 a_{n+3} - a_{n+4}) + \ldots \le 0$$ from the convexity condition.

This is from Apostol's book on Analysis:

Suppose $$a_n\searrow0$$ and $$s=\sum^\infty_{n=1}a_n$$, and $$s_n=\sum^n_{k=1}(-1)^{k+1}a_k$$. Then \begin{align} a_{n+1}-a_{n+2}&<\sum^\infty_{k=1}(a_{n+2k-1}-a_{n+2k})\\ &=(-1)^n(s-s_n)=a_{n+1}-\sum^\infty_{k=1}(a_{n+2k}-a_{n+2k+1})

Maybe this can be used to show the decay in errors that you are observing for $$a_n=\frac{1}{n^4}$$. Certainly $$|s-s_n| so at each step, the bounds improve.

• See xFioraMstr18's comment above: The error terms are not necessarily decreasing for an alternating series with decreasing absolute values. Sep 20, 2020 at 17:44
• @MartinR: do you have an example with strictly decreasing values ?
– user65203
Sep 20, 2020 at 18:16
• @YvesDaoust: Unfortunately not. I only cited a comment which I found relevant here, but I did not search myself for pathological examples. Sep 20, 2020 at 19:19