Prove the inequality following by induction

Question

So, I was stuck with this induction problem i.e

Show that P(n) : $\frac{(2n)!}{(n!)^2} > \frac{4^n}{n+1}$ for all integers n > 1

I tried doing the following and reached the answer but do not know if it is correct. Would appreciate input.

Basic Step: P(2) is true.

Inductive Step: For some k, P(k) is true.......

$\frac{(2n)!}{(n!)^2} > \frac{4^n}{n+1}$

P{k+1): is as follows:

We can show that $\frac{(2n)!}{(n!)^2} > \frac{4^n}{n+1} > \frac{4^n}{n+2}$

and

$\frac{(2(k+1)!}{((k+1)!)^2} =\frac{(2k+1)(2k+2)}{(k+1)^2}.\frac{(2k)!}{(k!)^2} > \frac{(2k+1)(2k+2)}{(k+1)^2}.\frac{4^k}{(k+2)}$

Now, for all integers $k > 0, 2k + 1 > 2k and 2k + 2 > 2k.$ Hence, we can replace this in the above formula

$\frac{(2(k+1)!}{((k+1)!)^2} > \frac{(2k)(2k)4^k}{(k+1)^2(k+2)}$

or

$\frac{(2(k+1)!}{((k+1)!)^2} > \frac{k^2 4.4^k}{(k+1)^2(k+2)}$

Now, since $k + 1 > k$, we have

$\frac{k^2}{(k+1)^2} < 1$

Hence, we replace and complete the proof:

$\frac{(2(k+1)!}{((k+1)!)^2} > \frac{4.4^k}{k+2}$

Welcome to Math SE. You conclusion that $\frac{k^2}{(k+1)^2} \gt 1$ is not true, e.g., for $k = 1$, we have $\frac{k^2}{(k+1)^2}=\frac{1}{4} \lt 1$. — John Omielan, Nov 21 '22 at 20:14
Does this answer your question? Prove that for $n \ge 2$ the follow inequality holds $\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}$. - found using an Approach0 search. There's also quite a few other duplicates, in the search result list and/or linked to various questions, e.g., ... — John Omielan, Nov 21 '22 at 20:23
(cont.) Prove by induction: $\frac{2^{2n}}{n+1}<\frac{(2n)!}{(n!)^2},n>1$, Problem proving inequality $\frac{4^n}{n+1}\leq \frac{(2n)!}{(n!)^2}$, Prove that $\frac {(2n)!} {(n!)^2} > \frac{4^n}{n + 1}$, etc. — John Omielan, Nov 21 '22 at 20:27
@JohnOmielan yeah a small typo, I meant the other way around i.e $\frac{k^2}{(k+1)^2}<1$ — Chaitanya Chaurasia, Nov 22 '22 at 20:38
Thank you for your response. However, note that with this correction, your final line of $\frac{(2(k+1)!}{((k+1)!)^2}\gt\frac{4.4^k}{k+2}$ doesn't necessarily follow from your previous line of $\frac{(2(k+1)!}{((k+1)!)^2}\gt\frac{k^2 4.4^k}{(k+1)^2(k+2)}$. — John Omielan, Nov 22 '22 at 20:43
@JohnOmielan Yeah my fault. I see it. Could you help me with the complete proof? I saw the other answers, but they are mostly the other way round, which is not the case for me. — Chaitanya Chaurasia, Nov 23 '22 at 01:41
Note many problem solution approaches are not successful. The inequality issue encountered with your particular approach means it doesn't work. — John Omielan, Nov 23 '22 at 02:57

Prove the inequality following by induction

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