Limit Sum

Taking n²⁰¹¹+ r ≈ n²⁰¹¹
Ans=1.

g(n) = \lim_{n\rightarrow \propto }\sum{} \frac{1}{\sqrt[2011]{n^{2011}+n}}

h(n) = \lim_{n\rightarrow \propto }\sum{} \frac{1}{\sqrt[2011]{n^{2011}+0}}

f(n) = \lim_{n\rightarrow \propto }\sum{} \frac{1}{\sqrt[2011]{n^{2011}+r}}

now

g(n)\leq f(n)\leq h(n)

it can easily be found the g(n)=h(n)=1

hence by sandwich theorem, f(n)=1

g(n) = \lim_{n\rightarrow \propto }\sum{} \frac{1}{\sqrt[2011]{n^{2011}+n}}

g(n) = \lim_{n\rightarrow \propto }\ \frac{n}{\sqrt[2011]{n^{2011}+n}}

dividing numerator and denominator by n ,

g(n) = \lim_{n\rightarrow \propto }\ \frac{1}{\sqrt[2011]{1+n^{-2011}}} =1

h(n) =1 in a similar way

still confused....:(
how does the numerator in your second step become n??

since it is a summation of n like terms. (note the summation sign)

ok ...i got that step.....

but what if i do the same thing directly in the question????
every term becomes zero

all the terms are infinitely small , but they are added infinitely many times to give a significant result.