Prove that $$1+ \frac{1}{2^3} + \cdot \cdot \cdot + \frac{1}{n^3} < \frac{5}{4} $$
I got no idea of how to approach this problem.
Asked
Active
Viewed 77 times
3
-
This is certainly not true for $n=1$. Without either telling us what $n$ is equal to or placing a quantifier in front of it, this inequality cannot be proven – 5xum Sep 18 '15 at 10:18
-
Yes, as @5xum said, you should first judge the problem then post it here. Here, obviously $n=1$ doesn't hold. So, you should place a quantifier. – Sep 18 '15 at 10:22
-
Should that be <? Otherwise, just hit the calculator till it is true? – Paul Sep 18 '15 at 10:29
-
What techniques are available to you? Integrals? Differentiation, Taylor's formula? Or only arithmetic and induction? – Daniel Fischer Sep 18 '15 at 10:37
-
Not Taylor's Formula. – exilednick Sep 18 '15 at 10:39
-
Are integrals known? – Daniel Fischer Sep 18 '15 at 10:40
-
I tried converting it into integral form but I was unable. – exilednick Sep 18 '15 at 10:43
3 Answers
3
For every $k \geqslant 2$ and $n \geqslant k$ we have
$$\sum_{m = k}^n \frac{1}{m^3} < \sum_{m = k}^n \int_{m-1}^m \frac{dt}{t^3} = \int_{k-1}^n \frac{dt}{t^3} < \int_{k-1}^\infty \frac{dt}{t^3}.$$
A suitable choice of $k$ yields the desired inequality.
Daniel Fischer
- 206,697
-
If $k=3$ ($k=2$ doesn't work), this gives $\sum_{m=3}^n\frac{1}{m^3}<\frac{1}{2(k-1)^2}=\frac{1}{8}$, so $\sum_{m=1}^n\frac{1}{m^3}<\frac{5}{4}$ – user236182 Sep 18 '15 at 11:00
-
3
Since: $$\begin{eqnarray*} \frac{1}{n^3} &=& \frac{1}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)+\frac{1}{2}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)+\frac{2n+1}{2n^3(n+1)^3}\\&\leq&\frac{1}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)+\frac{1}{2}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)+\frac{1}{4}\left(\frac{1}{n^4}-\frac{1}{(n+1)^4}\right)\tag{1}\end{eqnarray*}$$
$$ \zeta(3) \leq \frac{1}{2}+\frac{1}{2}+\frac{1}{4}=\frac{5}{4} $$
follows by creative telescoping. With an extra step we may also prove $\frac{6}{5}\leq\zeta(3)\leq\frac{5}{4}$.
More tight approximations (both upper and lower bounds) can be derived from the identity: $$ \zeta(3) = \sum_{n\geq 1}\frac{5(-1)^{n+1}}{2n^3\binom{2n}{n}}\tag{2}$$ that was crucial in Apery's proof of the irrationality of $\zeta(3)$.
Jack D'Aurizio
- 353,855
-
-
@GudsonChou: it is just a tag. It makes easier to comment some point of the proof (quite short, in this case). – Jack D'Aurizio Sep 18 '15 at 11:09
-
Oh, I see. I was under the impression that you created a tag but forgot to use it later on. – Yes Sep 18 '15 at 11:11
1
Another way to tackle these problems is by proving a stronger statement using induction.
For example, in this case you might prove that for $n\geq 2$ $$\sum_{j=1}^n\frac{1}{j^3}\leq \frac{5}{4}-\frac{4}{15n^2}.$$
The case $n=2$ is easily verified ($\frac{9}{8}\leq \frac{5}{4}-\frac{1}{21}$). Now for the induction step:
$$\sum_{j=1}^{n+1}\frac{1}{j^3}=\frac{1}{(n+1)^3}+\sum_{j=1}^{n}\frac{1}{j^3}\leq \frac{1}{(n+1)^3}+\frac{5}{4}-\frac{4}{15n^2}$$ and now what we have to verify is that $$\frac{1}{(n+1)^3}+\frac{5}{4}-\frac{4}{15n^2}\leq \frac{5}{4}-\frac{4}{15(n+1)^2}.$$
This is equivalent to $$\frac{1}{(n+1)^3}\leq \frac{4}{15}\cdot\frac{2n+1}{n^2(n+1)^2}$$ $$\frac{4}{15}\geq\frac{n^2}{(n+1)(2n+1)}$$ which is true because we have equality in $n=2$ and the function on the RHS is decreasing.
Actually, the most natural way to proceed is to write $S_n\leq \frac{5}{4}-\frac{A}{n^s}$ and to go the above steps to find $A$.
Sonner
- 3,108
-
I may be wrong, but I believe the sequence on the RHS is actually increasing. – user84413 Sep 19 '15 at 00:29