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}$

  • $\begingroup$ 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.) $\endgroup$
    – Mittens
    Sep 20, 2020 at 17:03

4 Answers 4


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 $m$.


Partial answer

$v$ 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.

  • $\begingroup$ Also that the errors are decreasing. $\endgroup$ Sep 20, 2020 at 15:50
  • $\begingroup$ @mathcounterexamples.net How does this answer my question? Do we have $n^{-4}\le e_n$ or something? $\endgroup$ Sep 20, 2020 at 15:54
  • $\begingroup$ You have $e_n < 1/(n+1)^4$. $\endgroup$ Sep 20, 2020 at 15:54
  • $\begingroup$ How do you get from $e_n<1/(n+1)^4$ to $\dots<e_{n+2}<e_{n+1}<e_n<\dots<e_1$? $\endgroup$ Sep 20, 2020 at 15:56
  • 2
    $\begingroup$ @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 $e_n=0.$ $\endgroup$ 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 $$\begin{align} a_{n+1}-a_{n+2}&<\sum^\infty_{k=1}(a_{n+2k-1}-a_{n+2k})\\ &=(-1)^n(s-s_n)=a_{n+1}-\sum^\infty_{k=1}(a_{n+2k}-a_{n+2k+1})<a_{n+1} \end{align} $$

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

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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .