Limit of two sequences 
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*$ \lim_{n\to \infty} \sqrt[n]{3^n+4^n} $ . I think the limit is $4$. I did :
$ \sqrt[n]{3^n+4^n} = 4 \sqrt[n]{(\frac{3}{4}) ^n+1}$ .Am I right?

*$ \lim_{n\to \infty} \frac{1}{1\cdot 4 } + \frac{1}{4\cdot 7} +...+\frac{1}{(3n-2)(3n+1)}  $. I know that for each $k$ , this sequence is the sum of $k$ terms, the smallest one is $ \frac{1}{(3n-2)(3n+1)}  $, and the largest is $ \frac{1}{1. 4 }$ . But when substituting and trying to use the squeeze thm, I get that the limit should be between $0$ and $\infty$, which gives me nothing.
Thanks in advance.

Hint:
For the second one use this fact that :
$$\frac{1}{(3n+1)(3n-2)}=\frac{-1}{3(3n+1)}+\frac{1}{3(3n-2)}$$ So the sum can be reduced to this one: $$S_n=\frac{1}3\left(\frac{1}{1}+\frac{1}{(3n+1)(3n-2)}\right)$$
