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.
On
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 $a_{n+1}=f(n,x)$.
On
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.
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