# prove that :$(1+\frac{1}{n})^k<1+\frac{k}{n}+\frac{k^2}{2n^2}$.

if $$n,k \in \mathbb{N^*}$$ and $$(k-1)^2

prove that :$$(1+\frac{1}{n})^k<1+\frac{k}{n}+\frac{k^2}{2n^2}$$.

my attempt:

$$(1+\frac{1}{n})^k+1=(1+\frac{1}{n}).(1+\frac{1}{n})^{k-1}+1\leq \sqrt{(1+\frac{1}{n})^2+1}\sqrt{((1+\frac{1}{n})^{(k-1)})^2+1}\leq \sqrt{(1+\frac{1}{n})^2+1}\sqrt{(1+\frac{1}{n})^{{(k-1)}^2}+1}$$

(because $$a^{n^2}\geq (a^n)^2$$for any positive integer $$n\geq 2,$$so what i did is true for $$(k-1\geq 2$$,and we can verifies the truth for $$k=1,2,3$$ ).

$$\leq \frac{(1+\frac{1}{n})^2+1+(1+\frac{1}{n})^{{(k-1)}^2}+1}{2}<\frac{(1+\frac{1}{n})^2+1+(1+\frac{1}{n})^{n}+1}{2}<\frac{1}{2}+\frac{1}{n}+\frac{1}{2n^2}+\frac{e+\epsilon }{2}+1$$(because $$\lim_{n \to +\infty }(1+\frac{1}{n})^{n}=e$$,so it is bounded by $$e+\epsilon$$).

$$<\frac{k}{n}+\frac{k^2}{2n^2}+\frac{e+\epsilon+1 }{2}+1$$.

so finally:$$(1+\frac{1}{n})^k<\frac{e+\epsilon+1 }{2}+\frac{k}{n}+\frac{k^2}{2n^2}$$.

i know the first term in the RHS that i get is maybe small but not than $$1$$.

this is the first method that i did (but i spend some time for it), i think the proof by recurrence maybe will be useful but want to prove it directly, without the proof by recurrence,

now i have two question :

1- does my attempt is true(i mean does my algebraic manipulations are true? )?

2- can you develop this method ?

The result is trivial if $$k=1$$ or $$k=2$$. Assume $$k>2$$. Note that the condition $$(k-1)^2 implies that $$(k-1)(k-2) By the binomial theorem, one needs to prove that $$\left(1+\frac 1 n\right)^k=\sum_{i=0}^k{k\choose i}\frac 1{n^i}<1+\frac kn+\frac {k^2}{2n^2},~~~~(1)$$ where the left hand side can be written as $$1+\frac kn+\frac {k(k-1)}2\cdot \frac 1{n^2}+\frac {k(k-1)(k-2)}{3!}\cdot \frac 1{n^3}+\cdots+\frac{k}{n^{k-1}}+\frac 1{n^k}$$
$$=1+\frac kn+\frac{k^2}{2n^2}-\frac k{2n^2}+\frac {k(k-1)(k-2)}{3!}\cdot \frac 1{n^3}+\cdots+\frac{k}{n^{k-1}}+\frac 1{n^k},$$ hence (1) is equivalent to showing that $$\frac k{2n^2}>\frac{k(k-1)(k-2)}{3!}\cdot\frac 1{n^3}+\cdots+\frac k{n^{k-1}}+\frac 1{n^k}.~~~~(2)$$ To prove (2), one inserts the following expression in the inequality, namely $$A:=\frac k{n^2}\cdot\left(\frac 1{3!}+\frac 1{4!}+\cdots+\frac 1{k!}\right)$$ and conducts termwise comparison. The inequality $$\frac k{2n^2}>A$$ is equivalent to $$\frac 12>\frac 1{3!}+\frac 1{4!}+\cdots+\frac 1{k!},$$ which follows directly from $$\frac 12>e-\left(1+1+\frac 1{2!}\right)\geq \frac 1{3!}+\frac 1{4!}+\cdots+\frac 1{k!}.$$ It remains to prove that $$A =\frac k{n^2}\cdot\left(\frac 1{3!}+\frac 1{4!}+\cdots+\frac 1{k!}\right) >\frac{k(k-1)(k-2)}{3!}\cdot\frac 1{n^3}+\cdots+\frac k{n^{k-1}}+\frac 1{n^k}.~~(3)$$ To do this, simply rewrite the right hand side as $$\frac k{n^2}\cdot\frac 1{3!}\cdot\frac{(k-1)(k-2)}n +\frac k{n^2}\cdot\frac 1{4!}\cdot \frac{(k-1)(k-2)}{n}\cdot\frac{k-3}n$$ $$+\frac k{n^2}\cdot \frac 1{5!}\cdot\frac {(k-1)(k-2)}n\cdot\frac{k-3}n\cdot\frac{k-4}n+\cdots+\frac k{n^2}\cdot\frac 1{k!}\cdot\frac{(k-1)(k-2)}n\cdot\frac{k-3}n\cdots\frac 1n.$$ Now by the beginning remark, the third factor and the factors onward of each summand are less than $$1,$$ hence inequality (3) holds. This completes the proof of inequality (2) by the insertion of $$A$$, hence the original inequality holds.
\begin{align} \left(1+\frac1n\right) \left(1+{k \over n}+{k^2\over 2n^2}\right) &=1+{k'\over n}+{k'^2\over 2n^2}\\ &+{(k'-1)^2\over 2n^3}-{1\over 2n^2} \end{align} where $$k'=k+1$$. The condition $$(k'-1)^2 applied to the right hand side gives $$\left(1+\frac1n\right)\times \left(1+{k \over n}+{k^2\over 2n^2}\right) <1+{k+1\over n}+{(k+1)^2\over 2n^2}.$$ By induction on $$k$$, $$\left(1+\frac1n\right)^k<1+{k\over n}+{k^2\over 2n^2}.$$ as long as $$(k-1)^2.