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I wanted to show $\left(1+\frac{x}{n}\right)^n\leq\left(1+\frac{x}{n+1}\right)^{n+1}$ using binomial expansion for $n\geq 2$ and $x\geq 0$.

So my idea was to expand both using binomial expansion and try to compare term-wise. The $k^{th}$ term of $\left(1+\frac{x}{n}\right)^n$ is $n\choose k $$\left(\frac{x}{n}\right)^k$, and similar for the other expansion the $k^{th}$ term is $n+1\choose k $$\left(\frac{x}{n+1}\right)^k$. I tried to compute the difference and tried to divide one by the other, neither yields any sufficient results. How should I approach this question?

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Let $f(x,n):=(1+x/n)^n$, then we need to show $f(x,n+1)\geq f(x,n)$ for $x\geq0,n\geq2$. Observe that the coefficient of $x^k$ in $f(x,n)$ is $$\frac{1}{n^k}\binom{n}{k}=\frac{1}{k!}\cdot\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\dotsm\frac{n-k+1}{n}.$$ Correspondingly the coefficient of $x^k$ in $f(x,n+1)$ is$$\frac{1}{(n+1)^k}\binom{n+1}{k}=\frac{1}{k!}\cdot\frac{n+1}{n+1}\frac{n}{n+1}\frac{n-1}{n+1}\dotsm\frac{n-k+2}{n+1}.$$ To solve the problem, all we need to do is to show that the latter is always greater than or equal to the former. To do so, we pair up each of the $k$ factors (excluding $1/k!$) by noting that $$\begin{split}\frac{n}{n}&=\frac{n+1}{n+1}\\ \frac{n-1}{n}&\leq\frac{n}{n+1}\\ \frac{n-2}{n}&\leq\frac{n-1}{n+1}\\ &\vdots\\ \frac{n-k+1}{n}&\leq\frac{n-k+2}{n+1} \end{split}$$ and multiplying all the inequalities together yields the result.

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Dividing is a good strategy. If you divide $n\choose k $$\left(\frac{x}{n}\right)^k$ by $n+1\choose k $$\left(\frac{x}{n+1}\right)^k$, you get $\frac{n-k+1}{n+1}\cdot\left(\frac{n+1}{n}\right)^k$. Now write $\frac{n-k+1}{n+1}$ as a telescoping product of $k$ terms of the form $\frac{m}{m+1}$, pair each up with a $\frac{n+1}{n}$ term, and use the fact that $\frac{m}{m+1}\leq\frac{n}{n+1}$ if $m\leq n$.

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