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How would I go about proving this?

$$ \displaystyle\sum_{r=1}^{n} \left( 1 + \dfrac{1}{2r} \right)^{2r} \leq n \displaystyle\sum_{r=0}^{n+1} \displaystyle\binom{n+1}{r} \left( \dfrac{1}{n+1} \right)^{r}$$

Thank you! I’ve tried so many things. I’ve tried finding a series I could compare one of the series to but nada, I tried to change the LHS to a geometric series but that didn’t work out, please could someone give me a little hint? Thank you!

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Let’s divide both parts by $n$. Then LHS becomes an arithmetic average of $\left(1+\frac{1}{2r}\right)^{ 2r }$, $r=1..n$. The RHS becomes (binomial coefficients) $\left(1+\frac{1}{n+1}\right)^{ n+1 }$. The function $x\to \left(1+\frac{1}{x}\right)^{ x }$ seems to be concave(need to check it, though), hence the average is not greater than the value in the middle of the segment.

Here is the proof that $(1+1/x)^x$ is concave for $x\ge 1$.

The second derivative of $(1+1/x)^x$ is $(1+1/x)^x$ times $$p(x)=\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}$$

Now for $x\ge 1$, we have $$\ln(1+1/x)-\frac{2}{1+x}\le \frac{1}{x}-\frac{2}{1+x}=\frac{1-x}{x(1+x)}\le 0$$ and $$\ln(1+1/x)\ge \frac{1}{x}-\frac{1}{2x^2}\ge 0,$$ so \begin{align*}p(x)&= \ln^2(1+1/x)-\frac{2\ln(1+1/x)}{1+x}+\frac{1}{(1+x)^2}-\frac{1}{x(1+x)^2}\\

&=\ln(1+1/x)(\ln(1+1/x)-2/(1+x))+\frac{x-1}{x(1+x)^2}\\

&\le \left(\frac{1}{x}-\frac{1}{2x^2}\right)\left(\frac{1}{x}-\frac{2}{1+x}\right)+\frac{x-1}{x(1+x)^2}\\

&=-\frac{(x-1)^2}{2x^3(1+x)^2}\le 0

\end{align*}

proving that $(1+1/x)^x$ is concave for $x\ge 1$.

And here’s the proof why $(1+1/x)^x$ is concave for $x\le 1$ (the case $x\ge 1$ is already proven by @TCL).

We study the second derivative with respect to $x$, it is equal to (thanks to TCL) to

$\displaystyle \left(1+ \frac 1x\right)^x\left(\left(\ln (1+1/x)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}\right)$;

First of all, $\ln (1+1/x)\ge\frac{1}{1+x}$; we say that $y=1/x\ge 1$ and then study $(\ln (1+y)-\frac{y}{1+y})$ – this function is positive in $y=1$ and it’s derivative with respect to $y$ is positive for $y\ge 1$, hence the function itself is always positive; therefore $\sqrt{\left( \ln (1+1/x)-\frac{1}{1+x}\right)^2}= \ln (1+1/x)-\frac{1}{1+x}$.

Now we study the sign of $\displaystyle \left(\left(\ln (1+1/x)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}\right)$; it’s equivalent to study the sign of

$\displaystyle \ln (1+1/x)-\frac{1}{1+x} -\frac{1}{\sqrt x(1+x) } $. Once again, we change $y=1/x\ge 1$:

$\displaystyle \ln (1+y)-\frac{y}{1+y} -\frac{y\sqrt y}{ (1+y) } $.

In $y=1$ the last expression is negative (we have concavity for $x\ge 1$, after all); its derivative for $y\ge 1$ is (after several manipulations)

$\displaystyle 0.5(1+y)^{-2}(2y -3\sqrt y – y^{3/2})\le 0$,

which ensures that $ \ln (1+1/x)-\frac{1}{1+x} -\frac{1}{\sqrt x(1+x) } $ remains negative for $y\ge 1$; therefore, $(1+1/x)^x$ is concave for $x\le 1$. QED

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