Limit of two sequences 
*

*$ \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.
 A: 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: $$S_n=\frac{1}3\left(\frac{1}{1}+\frac{1}{(3n+1)(3n-2)}\right)$$
A: 1- Your first limit is correct.
2- As I suggested in the comments, you have to partialize the fractions into two, as such:
$$\frac{1}{(3n+1)(3n-2)}= \frac{1}{3} \left(\dfrac{-1}{(3n+1)}+\dfrac{1}{(3n-2)}\right)$$
Using the fact that it is a telescoping sum, you now have to find the:
$$\sum_{n=1}^{\infty}\left(\frac{-1}{(3n+1)}+\frac{1}{(3n-2)}\right)$$
Listing out the first few terms and the $n$-th term the cancellations become clear,
$$\lim \limits_{n \to \infty} \left[\frac{-1}{4} +\frac{-1}{7}+\frac{-1}{10}...\frac{-1}{3n-2} +\frac{-1}{3n+1} \right] + \left[ 1 + \frac{1}{4} + \frac{1}{7}+...\frac{1}{3n-2} \right]$$
$$=\lim \limits_{n \to \infty} \left(1-\frac{1}{3n+1}\right)=1$$
Therefore, the final sum is $\dfrac{1}{3}$
A: For every $n$
$$
4^n<3^n+4^n< 2\cdot 4^n,
$$
thus
$$
4<\sqrt[n]{3^n+4^n}< 2^{1/n}\cdot 4.
$$
