Show that any sequence of positive numbers satisfying $0\lt  \frac{a_{n+1}}{a_n} \leq 1 + \frac{1}{n^2}$ must converge 
Show that any sequence of positive numbers $(a_n)$ satisfying $$0< \frac{a_{n+1}}{a_n} \leq 1+ \frac{1}{n^2}$$
  must converge.

I have tried taking the limit of the inequality which yields that $0 \leq \lim \frac{a_{n+1}}{a_n} \leq 1$. If  $\lim \frac{a_{n+1}}{a_n} \lt 1$, then by the ratio test we have $\sum a_n$ converges, thus  $a_n \to 0$. I am trying in particular trying to show that if $\frac{a_{n+1}} {a_n} \to 1$, then $a_n$ is Cauchy thus convergent, of which I am having some trouble.
 A: Let $L = \liminf_{m \rightarrow \infty} a_m$. This will actually be the limit. We first show $L$ is finite. To see this, multiply ${a_{n+1} \over a_n} < 1 + {1 \over n^2} < e^{1 \over n^2}$ together for $1 \leq n < m$. You then get 
$$a_m \leq e^{\sum_{n=1}^{m-1}{1 \over n^2}}a_1$$
$$< e^{\pi^2 \over 6}a_1$$
Hence the sequence is bounded and $L$ is finite.
Let $\epsilon > 0$. Since $\liminf_{m \rightarrow \infty} a_m = \sup_n \inf_{m \geq n} a_m$, for each $n, \inf_{m \geq n} a_n \leq L$ and we can choose $m$ arbitrarily large such that $a_m \leq L + \epsilon$. Suppose $M > m$.
Then multiplying ${a_{n+1} \over a_n} < 1 + {1 \over n^2} < e^{1 \over n^2}$ together for $M > n \geq m$, you get 
$$a_M < e^{\sum_{n=m}^{M-1}{1 \over n^2}}a_m$$
$$< e^{\sum_{n=m}^{\infty}{1 \over n^2}}a_m$$
Given $\epsilon > 0$, if $m$ is large enough the sum $\sum_{n=m}^{\infty}{1 \over n^2}$ can be made less than $\epsilon$. Thus for any $M > m$ you have
$$a_M < (L + \epsilon) e^{\epsilon}$$
Since $\epsilon$ was arbitrary you therefore have
$\limsup_{m \rightarrow \infty}a_m \leq L$.
Since $L$ was defined as $\liminf_{m \rightarrow \infty} a_m$, the overall limit exists and equals $L$. 
A: Hint: For $m<n$, find a reasonably good estimate for $\dfrac{a_n}{a_m}$. Use this estimate to show that our sequence is Cauchy.  
Details:  For fun we give an elementary approach to the estimates. Note that if $k \ge 2$, then
$$1+\frac{1}{k^2}<\frac{1}{1-\frac{1}{k^2}}=\frac{k^2}{(k-1)(k+1)}.$$
Let $2\le m<n$.  Then
$$\prod_{k=m}^n \left(1+\frac{1}{k^2}\right)< \prod_{k=m}^n \frac{k^2}{(k-1)(k+1)}.$$
Write down the product of the first few terms on the right-hand side, and observe the beautiful cancellations. A little examination shows that 
$$\prod_{k=m}^n \frac{k^2}{(k-1)(k+1)}=\frac{mn}{(m-1)(n+1)}<1+\frac{1}{m-1}.\qquad (\ast)$$
Now we complete the proof of convergence. From the inequality in  the statement of the problem, we conclude using $(\ast)$ that 
$$0<\frac{a_n}{a_m}<1+\frac{1}{m-1}, \qquad\text{and therefore}\qquad |a_n-a_m| <\frac{a_m}{m-1}. $$
But from $(\ast)$ we can see that $a_m<2a_2$.  So by picking $m$ large enough, and $n>m$, we can make $|a_n-a_m|<\epsilon$ for any preassigned $\epsilon$, that is, the sequence $(a_n)$ is Cauchy. 
A: Hint: Define
$$
b_n=\prod_{k=1}^{n-1}\left(1+\frac{1}{k^2}\right)
$$
and note that $\displaystyle\frac{b_{n+1}}{b_n}=1+\frac{1}{n^2}$. Show that $b_n\le e^{\pi^2/6}$ for all $n$. What does this say about $\lim\limits_{n\to\infty}b_n$?
Consider the sequence $c_n=\dfrac{a_n}{b_n}$. What can you say about $\dfrac{c_{n+1}}{c_n}$?
Further Explanation of Hint:
$$
\begin{align}
\frac{c_{n+1}}{c_n}
&=\frac{a_{n+1}}{a_n}\left/\frac{b_{n+1}}{b_n}\right.\\
&=\frac{a_{n+1}}{a_n}\left/\left(1+\frac{1}{n^2}\right)\right.\\
&\le1
\end{align}
$$
So $c_n$ is non-increasing and positive.
