What is the simplest way to prove this inequality without calculator and without calculus (I don't know calculus):

$$\frac12-\frac13+\frac14-\frac15+...+\frac1{2018}-\frac1{2019} \in \left(\frac14, \frac13\right)$$

For numbers $> \frac{1}{4}$, I can prove like this:

$$\frac12-\frac13+\frac14-\frac15+...+\frac1{2018}-\frac1{2019} > \frac12-\frac13+\frac14-\frac15+\frac16-\frac17+\frac18-\frac19=$$

$$\frac16+\frac1{20}+\frac1{42}+\frac1{72} = 0.254... > \frac{1}{4}$$

but I don't like it. Are there cleaner ways?

Edit: The original question is:

$$\frac12-\frac13+\frac14-\frac15+...+\frac1{2002}-\frac1{2003} \in \left(\frac14, \frac13\right)$$

from Challenging Problems in Algebra by Charles Salkind.

  • $\begingroup$ @amWhy, I have a version with $...+\frac1{2002}-\frac1{2003}$ from an older book. I changed it to keep up with the times after checking it is true by computer. Is that ok? $\endgroup$
    – user750196
    Commented Feb 13, 2020 at 19:19
  • $\begingroup$ That's fine, @ryan. But it would still be good to cite the source of the original problem, albeit modified. $\endgroup$
    – amWhy
    Commented Feb 13, 2020 at 19:20
  • $\begingroup$ @amWhy, I edited. $\endgroup$
    – user750196
    Commented Feb 13, 2020 at 19:27
  • $\begingroup$ Thanks, @ryan! ${}$ $\endgroup$
    – amWhy
    Commented Feb 13, 2020 at 19:28
  • 2
    $\begingroup$ The sums in which you sum up to an even term (I am counting starting from zero) are decreasing, while those up to an odd term are increasing. The distance between a sum to an even term and the sum up to the next odd term is bounded by the absolute value of the next term added. If you sum up arbitrarily number of terms the sum gets closer to $1-\log(2)$. The distance of a partial sum to this value is bounded by the absolute value of the last term added. Therefore, it is enough to check that $1-\log(2)\in(1/4,1/3)$ and that $1/2019$ is not larger than $|1-\log(2)-1/3|$ and $|1-\log(2)-1/4|$ $\endgroup$
    – user748968
    Commented Feb 13, 2020 at 19:31

4 Answers 4


Without using calculus, one may observe that $$S_n:=\frac12-\frac13+\frac14-\frac15\pm\ldots +\frac{(-1)^{n+1}}n $$ has the following properties:

If $n$ is even, then $S_{n+2}<S_n$. If $n$ is odd, then $S_{n+2}>S_n$.

Both follow immediately from $\frac1{n+1}-\frac1{n+2}=\frac1{(n+1)(n+2)}>0$. By this, if $a,b<n$, $a$ odd, $b$ even, then $S_a<S_n<S_b$. By checking, the smallest useful $a,b$ are $a=9$ (as you also found) and $b=20$ (with $S_{20}=0.331\ldots <\frac13$). If you dislike computing $S_9$, I can only imagine what you think about computing $S_{20}$ without aid!

For a better approach, we need, still avoiding calculus, a better estimate for for $S_n-S_N$ when $N\gg n$. We invest some manageable algebraic work in order to avoid much numerical computation. Revisiting the above result, we have for odd $n\ge k$ $$\begin{align}S_{n+2}-S_n&=\frac1{n+1}-\frac1{n+ 2}\\&=\frac1{(n+1)(n+2)}\\&\ge \frac {\frac k{k+1}}{n(n+2)}\\&=\frac k{2(k+1)}\left(\frac1n-\frac1{n+2}\right). \end{align}$$ Summing this for $n=k, k+2, k+4, \ldots$, we obtain a telescoping sum and thereby $$ \tag1\begin{align}S_N-S_k&\ge \frac k{2(k+1)}\left(\frac1k-\frac1N\right)\\&=\frac1{2(k+1)}-\frac{k}{2(k+1)N}\end{align}$$ for $N>k$ and both odd. The same argument leads to $$ \tag2S_k-S_N\ge \frac1{2(k+1)}-\frac{k}{2(k+1)N}$$ for $N>k$ and both even.

For $N=2019>9$ and $k=3$, $(1)$ leads to $$S_N\ge \left(\frac12-\frac13\right)+\frac18- \frac 3{8N}=\frac14+\frac1{24}- \frac 3{8N}>\frac14.$$ For $N=2018>24$ and $k=4$, $(2)$ leads to $$S_N\le \left(\frac12-\frac13+\frac14\right)-\frac1{10}+\frac 4{10N} =\frac13-\frac1{60}+\frac 4{10N}<\frac13.$$ Therefore, $$ \frac14<S_{2019}<S_{2018}<\frac13.$$

  • $\begingroup$ I think this is very clever (+1). $\endgroup$
    – LHF
    Commented Feb 13, 2020 at 21:16

First, this is (almost) a truncated Alternating Series. So the infinite result is:

$$ \sum_{n=0}^{\infty} \frac{-(-x)^n}{n} = \ln(1+x) $$

Thus for $x=1$, we have $\ln(2)=0.693147$ as the end result. And you started at $+1/2$,

Hence your series converges to $1-\ln(2)=0.306852$. Now, the typical epsilon-delta definition of convergence tells you that for any given $\varepsilon$ there will be $n_0$ such that the series being truncated at any term $n \gt n_0$ implies the result to be within $\varepsilon$ of the limit value. So pick $\varepsilon=0.3$.

Now, notice that THEOREM 5.14 in this reference tells us that the error in such an alternating series is no larger than the the first excluded term, which in your case is $1/2020=0.000495$. So yeah, you're well within the interval given.



$$S = \frac{1}{2}-\frac{1}{3}+\ldots +\frac{1}{2018}-\frac{1}{2019}$$

For the right side, I will use the following identity (Botez-Catalan), which can be proven with induction:

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{2n} = \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}$$

We have:


and using Cauchy-Schwarz:

$$ \begin{aligned} \frac{1}{1010}+\frac{1}{1011}+\ldots+\frac{1}{2019}&\geq \frac{(1+1+\ldots+1)^2}{1010+1011+\ldots+2019}\\ &=\frac{1010^2}{505(2\cdot 2019-1009)}\\ &=\frac{2020}{3029}>\frac{2020}{3030}=\frac{2}{3} \end{aligned} $$

Thus, we have $S < \dfrac{1}{3}$. For the left side, I couldn't come up with a better approach than the OP.


For the first part, I suggest this slight modification of the proof in the OP: $$ \frac1{2\cdot3}+\frac1{4\cdot5}+\frac1{6\cdot7}+\frac1{8\cdot9} > \frac1{2\cdot3}+\frac1{4\cdot5}+\frac1{6\cdot8}+\frac1{8\cdot10} = \frac{40 + 12 + 5 + 3}{240} = \frac{60}{240} = \frac14, $$ which is surely clean enough.

Next, for any $n \geqslant 2$: \begin{align*} \frac14 - \frac15 + \frac16 - \frac17 + \cdots + \frac1{2n} - \frac1{2n+1} & = \frac1{4\cdot5} + \frac1{6\cdot7} + \cdots + \frac1{2n(2n+1)} \\ & < \frac1{3\cdot5} + \frac1{5\cdot7} + \cdots + \frac1{(2n-1)(2n+1)} \\ & = \frac12\left(\frac13 - \frac15 + \frac15 - \frac17 + \cdots\ + \frac1{2n-1} - \frac1{2n+1}\right) \\ & = \frac12\left(\frac13 - \frac1{2n+1}\right) \\ & < \frac16, \end{align*} which proves the second part.

  • $\begingroup$ I very much like this solution. If I could, I would accept two answers. $\endgroup$
    – user750196
    Commented Feb 14, 2020 at 21:09

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