Question

Question #e5cde

Answer 1

Your sum can be expressed as a Riemann Sum. Hence

`\frac 1{n+1}+...+\frac 1{2n}=\int_n^{2n}\frac {1}{x}dx=ln(2n)-ln(n)=ln(2)~~0.693`

Also notice that

`1/2=\frac n{2n}\le\frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}\le\frac n{n+1}`

As well as

`\sum_{k=1}^n\frac1{n+k}\leq 1 - \sum_{k=1}^n\frac{k}{2n^2}=1-\frac{n(n+1)}{4n^2}=3/4-\frac1{4n}`

because

`\frac1{n+k}\leq\frac1n-\frac{k}{2n^2}`

Answer 2

The limit is `ln2`, but I don't see how to get it using the sandwich theorem.

Explanation:

We want `lim_(nrarroo)sum_(i=1)^n1/(n+i)`

`lim_(nrarroo)sum_(i=1)^n1/(1+(i/n))1/n`

This is an integral.

`int_a^b f(x) dx = lim_(nrarroo)sum_(i=1)^n f(x_i) Delta x`

Where `Delta x = (b-a)/n` `" "` and `" "` `x_i = a+iDeltax`

Here we can have `Delta x = 1/n`
and use either

`a=1` so `b= 2` and `x_i = 1+i/n`
In which case `f(x) = 1/x` and we have `int_1^2 1/x dx = ln2`

Or we can use

`a=0` so `b=1` and `x_i = i/n`
In which case `f(x) = 1/(1+x)` and we have `int_0^1 1/(1+x) dx = ln2`