How to obtain this upper bound for a finite sum?

Question

An exercises asks me to show that $$\sum_{k = 1}^n \frac{1}{k^3} \leq \frac{3}{2} + \frac{1}{2n}$$

Now I tried, but I cannot obtain that specific result. This is what I tried, and honestly it's all wrong since I know that as $n\to +\infty$ the series converge to $\zeta(3)$, while with my method the series diverges when $n \to +\infty$:

$$\frac{1}{k^3} = 1 + \frac{1}{8} + \frac{1}{3^3} + \frac{1}{4^3} + \frac{1}{5^3} + \ldots \leq 1 + \frac{1}{8} + \frac{1}{3^3} + \frac{1}{3^3} + \frac{1}{3^3} + \ldots = \frac{9}{8} + \frac{n-2}{27}$$

I tried to turn things into some geometric series, but I couldn't find a proper bound. For example $k^3 \geq 2^k$ but this fails when $k \geq 10$, so I cannot turn this into $\frac{1}{k^3} \leq \frac{1}{2^k}$.

Can you give me some hint please?

No induction method.

Answer 1

I can do better.

For $k>1,$ $$\frac1{k^3}<\frac1{k(k^2-1)}=\frac12\left(\frac1{k(k-1)}-\frac1{k(k+1)}\right)$$

So the right side becomes a telescoping sum. If $f(k)=\frac12\cdot\frac1{k(k-1)},$ we get:

$$\sum_{k=1}^n\frac1{k^3}\leq 1+\sum_2^n (f(k)-f(k+1))=1+f(2)-f(n+1)=\frac54-\frac1{2n(n+1)}$$

Answer 2

We prove a stronger inequality:

For $k>1$ we have $$\frac{1}{k^3}<\frac{1}{(k+1)k(k-1)}$$ so \begin{align}\sum_{k=1}^n\frac{1}{k^3} &\le1+\sum_{k=2}^n\frac{1}{(k+1)k(k-1)}\\ &=1+\frac{1}{2}\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}+\frac{1}{k+1}-\frac{1}{k}\right)\\ &=1+\frac{1}{2}\left(1-\frac{1}{n}+\frac{1}{n+1}-\frac{1}{2}\right)\\ &=\frac{5}{4}+\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n}\right) \end{align}

Answer 3

You can show that $a_k=k^{-3}$ is a positive, decreasing sequence. Then $$\sum_{k=1}^n k^{-3}\leq (1)^{-3}+\int_1^n x^{-3}\,dx$$

Integrating, we get $-\dfrac{1}{2n^2}+\dfrac{1}{2}$ so that the right-hand side above is equal to $\dfrac{3}{2}-\dfrac{1}{2n^2}$. The resulting inequality is strictly better than the one asked for in the problem statement.

Answer 4

Here is a more complicated way to prove an upper bound close to $\frac32$.

For all $x$, we have $e^x \ge 1+x$. So $e^{-x} \ge 1-x$ and $e^x \le \frac1{1-x}$. In particular, reserving the first term of our sum for later, we have $$ \exp\left(\sum_{k=2}^n \frac1{k^3}\right) = \prod_{k=2}^n e^{1/k^3} \le \prod_{k=2}^n \left(\frac1{1-1/k^3}\right) = \prod_{k=2}^n \left(\frac{k^3}{k^3-1}\right). $$ To get an upper bound, let's take $\frac{k^3}{k^3-1} \le \frac{k^3+1}{k^3-1} = \frac{k+1}{k-1} \cdot \frac{k^2-k+1}{k^2+k+1}$, because then the product telescopes: $$ \prod_{k=2}^n \frac{k^3+1}{k^3-1} = \prod_{k=2}^n \frac{k+1}{k-1} \cdot \prod_{k=2}^n \frac{k^2-k+1}{k^2+k+1} = \frac{n(n+1)}{2} \cdot \frac{3}{n^2+n+1}. $$ We conclude that $$\sum_{k=2}^n \frac1{k^3} \le \ln \prod_{k=2}^n \frac{k^3+1}{k^3-1} = \ln \frac32 + \ln\left(\frac{n^2+n}{n^2+n+1}\right) \le \frac12 - \frac1{n^2+n+1}$$ where in the last step we apply the inequality $\ln(1+x) \le x$ twice (another variant of $e^x \ge 1+x$).

We are now ready to put the $k=1$ term back in; we couldn't handle it before, because $\frac{k^3}{k^3-1}$ doesn't behave nicely when $k=1$. This brings us to our final conclusion that $$\sum_{k=1}^n \frac1{k^3} \le \frac32 - \frac1{n^2+n+1}.$$