$\displaystyle \lim_{x\rightarrow\infty}\left (\frac{1+2^{199}+3^{199}+4^{199}...+n^{199}}{n^{20 0}}\right )$

Is equal to:

a. 1/201

b.1/200

c.1/199

d.1/98

e. None of these

My attempt:

I have no clue how to take this limit so I just tried to use calculator guessing.

$\displaystyle \frac{1}{1^{200}}=1$

$\displaystyle \frac{1+2^{199}}{2^{200}}=.5$

$\displaystyle \frac{1+2^{199}+3^{199}}{3^{200}}=.3333...$

$\displaystyle \frac{1+2^{199}+3^{199}+4^{199}}{4^{200}}=.25...$

And my calculator wont go past this so I just figured that it continued with the pattern that it has now.

so

keeping with the pattern:

if n=1

it's

1^{-1}

if n=2

it's

2^{-1}

if n=3

it's

3^{-1} ect.

it must be zero because 1/infinity=0

so e

The answer however is B. Does anyone know how to work this limit?