Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ I am just trying to calculate 
$$\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}.$$
To do this I apply formula for sum of fourth powers of $n$ number. My result: $$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}=1$$. I'm intrested in finding other method to solve the following problem.  

If
$a_n > 0,
s_n = \sum_{k=1}^n a_n,
r_n = s_{n+1}/s_n$,
then,
if $a_{n+1}/a_n \to 1$,
$r_n \to 1$.
Proof:
$r_n-1
=a_{n+1}/s_n
$.
Since 
$a_{n+1}/a_n \to 1$,
then,
for any fixed $m$,
$a_{n+k}/a_n \to 1$
for $1 \le k \le m$.
Therefore,
for any fixed $m$,
$\dfrac{a_{n+1}}{s_n}
\lt \dfrac{a_{n+1}}{\sum_{k=0}^{m-1} a_{n-k}}
\to \dfrac1{m}
$
so that
$r_n-1
=\dfrac{a_{n+1}}{s_n}
\to 0
$.
Your case is
$a_n = n^4$,
and it is easy to show that
$a_{n+1}/a_n \to 0$.
This holds for
$a_n = n^p$
for any fixed $p$.
