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Let $$f(k)=\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{\infty}\dfrac{1}{t^2}.$$ Prove $$f(k+1)>f(k).$$

My idea:

$$f(k+1)-f(k)=\dfrac{(k+2)^{k+2}}{(k+1)^{k+1}}\sum_{t=k+2}^{\infty}\dfrac{1}{t^2}-\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{\infty}\dfrac{1}{t^2}$$ and following is very ugly. I think someone can use nice methods. Thank you.

This problem is from this topic because $$f(k)=\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{\infty}\dfrac{1}{t^2}<\dfrac{(k+1)^{k+1}}{k^k}\left(\dfrac{1}{k}-\dfrac{1}{n}\right)$$ and note that $$\lim_{k\to\infty}\dfrac{(k+1)^{k+1}}{k^k}\dfrac{1}{k}=e.$$

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math110
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2 Answers2

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It seems the following.

my idea:

$$f(k+1)-f(k)=\dfrac{(k+2)^{k+2}}{(k+1)^{k+1}}\sum_{t=k+2}^{\infty}\dfrac{1}{t^2}-\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{\infty}\dfrac{1}{t^2}$$ and following is very ugly. I think someone can use nice methods

Put $S(k)=\sum_{t=k+2}^{\infty}\dfrac{1}{t^2}$. We have to show that

$$0< f(k+1)-f(k)=\dfrac{(k+2)^{k+2}}{(k+1)^{k+1}}S(k)-\dfrac{(k+1)^{k+1}}{k^k}\left(S(k)+\dfrac{1}{(k+1)^2}\right)$$

or

$$\dfrac 1{(k+1)^2}<S(k)\left(\dfrac{(k+2)^{k+2}k^k}{(k+1)^{2k+2}}-1\right).$$

Now we estimate from below the value $X(k)=\dfrac{(k+2)^{k+2}k^k}{(k+1)^{2k+2}}$.

$$X(k)=\left(1-\frac 1{(k+1)^2}\right)^k\left(\dfrac {k+2}{k+1} \right)^2.$$

By strengthened Bernoulli inequality, which can be easily proved by induction, we have $$(1-x)^k\ge 1-kx+\frac {k(k-1)x^2}{2}-\frac{k(k-1)(k-2)x^3}{6},$$

for each integer $k\ge 0$ and real $x\ge -1$. Therefore

$$X(k)\ge Y(k)=\left(1-\frac {k}{(k+1)^2}+\frac {k(k-1)}{2(k+1)^4}-\frac{k(k-1)(k-2)}{6(k+1)^6}\right)\left(\dfrac {k+2}{k+1} \right)^2$$

and $$Y(k)-1=\frac{6k^7+45k^6+146k^5+266k^4+297k^3+202k^2+76k+18}{6(k+1)^8}>0.$$

Therefore it suffices to show that

$$S(k)>Z(k)=\frac 1{(k+1)^2(Y(k)-1)}=$$ $$\frac{6(k+1)^6}{6k^7+45k^6+146k^5+266k^4+297k^3+202k^2+76k+18}.$$

Now we estimate $S(k)$. For this purpose we consider the sum $$\frac 1{k+\frac 32}=S_1(k)=\sum_{t=k+2}^{\infty}\left(\dfrac{1}{t-\frac 12}-\dfrac{1}{t+\frac 12}\right).$$

Then

$$0\le S_1(k)-S(k)=\sum_{t=k+2}^{\infty}\left(\dfrac{1}{t^2-\frac 14}-\dfrac{1}{t^2}\right) =\frac 14\sum_{t=k+2}^{\infty}\dfrac{1}{t^4-\frac {t^2}4}.$$

Consider the sum

$$\frac 1{\left(k+\frac 32\right)^3}=S_2(k)=\sum_{t=k+2}^{\infty}\left(\dfrac{1}{\left(t-\frac 12\right)^3}- \dfrac{1}{\left(t+\frac 12\right)^3}\right)=$$ $$=\sum_{t=k+2}^{\infty}\dfrac{3t^2+\frac 14}{\left(t^2-\frac 14\right)^3}\ge 3\sum_{t=k+2}^{\infty}\dfrac{1}{t^4-\frac {t^2}4}=12\left(S_1(k)-S(k)\right).$$

Therefore

$$S(k)\ge S_1(k)-\frac{S_2(k)}{12}=\frac 1{k+\frac 32}-\frac 1{12\left(k+\frac 32\right)^3}.$$

Now the direct check shows that

$$S(k)-Z(k)\ge S_1(k)-\frac{S_2(k)}{12}-Z(k)=$$ $$\frac 23\cdot\frac{18k^7+159k^6+628k^5+1375k^4+1740k^3+1319k^2+680k+225} {(6k^7+45k^6+146k^5+266k^4+297k^3+202k^2+76k+18)(2k+3)^3}>0.\square$$

Alex Ravsky
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First, recall the series appear in $f(k)$ has an integral representation: $$\sum_{s=1}^{\infty}\frac{1}{(k+s)^2} = \int_{0}^{\infty}\frac{t e^{-kt}}{e^t-1}dt =\int_{0}^{\infty}\frac{2t e^{-2kt}}{e^{2t}-1}d2t =4\int_0^{\infty}\frac{t}{e^t-e^{-t}} e^{-(2k+1)t} dt $$ Consider the integral $I(\eta)$ defined below and integrate it by parts:

$$\begin{align} I(\eta) =& \int_{0}^{\infty}\frac{\eta t}{e^t - e^{-t}} e^{-\eta t}dt = \int_0^{\infty}\frac{t}{e^t - e^{-t}}d(1 - e^{-\eta t})\\ =& \left[\frac{t}{e^t - e^{-t}}(1 - e^{-\eta t})\right]_0^{\infty} - \int_0^{\infty}(1-e^{-\eta t}) d (\frac{t}{e^t - e^{-t}})\\ =&\int_0^{\infty}(1-e^{-\eta t}) \varphi(t) dt \end{align}$$ where $\varphi(t) = (\frac{-t}{e^t - e^{-t}})' = \frac{t(e^t+e^{-t})}{(e^t-e^{-t})^2} - \frac{1}{e^t-e^{-t}}$.

The integrand of $I(\eta)$ has two factors. If one fix $t$ and consider the first factor $1 - e^{-\eta t}$ as a function of $\eta$, it is strictly increasing. For the second factor $\varphi(t)$, it is easy to check $\varphi(t) > 0$ for all $t > 0$. Combine these, we can conclude $I(\eta)$ is a increasing function in $\eta$.

As a consequence, we obtain:

$$(k + \frac12)\sum_{t=k+1}^{\infty}\frac{1}{t^2} = 2I(2k+1) \le 2I(2k+3) = (k + \frac32)\sum_{t=k+2}^{\infty}\frac{1}{t^2} \tag{*1}$$

Second, consider the function:

$$\psi(x) := (x+\frac12)\log(x+\frac12) - (x-\frac12)\log(x-\frac12)-\log x$$

We have: $$\begin{align} \psi'(x) &= \log(x+\frac12)-\log(x-\frac12)-\frac{1}{x} = \log\left(\frac{1+\frac{1}{2x}}{1-\frac{1}{2x}}\right) - \frac{1}{x}\\ \psi''(x) &= -\frac{1}{x^2(4x^2-1)} \end{align}$$ What is sort of obvious is $\psi'(x) \to 0$ as $x \to \infty$ and $\psi''(x) < 0$ for $x > \frac12$. From this we can conclude $\psi'(x) > 0$ for $x \in (\frac12,\infty)$ and $\psi(x)$ is an increasing function there. As a consequence, we get:

$$\frac{(k+1)^{k+1}}{k^k (k+\frac12)} = e^{\psi(k+\frac12)} \le e^{\psi(k+\frac32)} = \frac{(k+2)^{k+2}}{(k+1)^{k+1}(k+\frac32)}\tag{*2}$$

Multiply $(*1)$ with $(*2)$ gives us $f(k) \le f(k+1)$ immediately. `

achille hui
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