Prove convergence and find the value of the limit of the sequence Sequence
$$a_{n+1}=(1+\frac{1}{3^n})a_n$$
$$a_1=1$$
The question asks to prove its convergence and find its limit. I have tried all the usual ways but am unable to solve it. The question also says that we should try to prove that it is bounded by 3. Please help.
 A: Hint:
Let $b_n=\ln a_n$, then $b_n=\sum_{k=1}^{n-1}\ln (1+\frac{1}{3^k})$.
Since $\ln (1+x)<x$, so $b_n<\sum_{k=1}^{n-1} \frac{1}{3^k}$ converges.
For the exact value, I do not think this limit can be computed. see wolframalpha
A: For the convergence, see the answer of Ma Ming.

Consider:
$$f(n,x)=\prod_{k=1}^n(1+x^k)$$
We are interested in the asymptotical behavior of $f(n,x)$ for $n\to\infty$ with $0\le x<1$. Notice that only the first $m$ terms in the product can contribute to terms of order $x^m$. Every occurring $x^m$ corresponds to a partition of $m$ of the form $\lambda_1>\lambda_2>\ldots>\lambda_\ell>0$, $\lambda_1+\ldots+\lambda_\ell=m$ through $x^m=x^{\lambda_1}\ldots x^{\lambda_\ell}$. Let $p(m)$ be the number of such partitions of $m$. This shows that:
$$\bigg[f(n,x)\bigg]_{x^m}=\tilde{p}(m)$$
if $n\ge m$. Thus we have:
$$f(n,x)=\sum_{k=0}^n\tilde{p}(k)x^k+o(x^{n+1})$$
In the limit $n\to\infty$ you get the generating function of restricted partitions:
$$f(x)=\prod_{k=1}^\infty(1+x^k)=\sum_{k=0}^\infty\tilde{p}(k)x^k$$
This doesn't have any closed form that I know about, and I don't think it can be evaluated for specific $x>0$ (except numerically, of course).
Note: Obviously the case that interests us is $x=\frac{1}{3}$, for which we have $a_{n+1}=f(n,x)$.
A: As another answer alluded, you can bound you sequence above and thus show that it converges. As for the exact value, it is equal to the series $\sum_n a_n (1/3)^n$ where $a_n$ counts the number of ways to choose any number of distinct positive integers that add up to be $n$ (and $a_0 = 1$). This seems like a hard series value to calculate. It's possible that no closed-form formula for the limit of your sequence can be found.
