Question

Evaluate the following limit.

Answer 1

\[\lim_{n \rightarrow \infty} \sum_{k=n+1}^{2n} \frac{ 1 }{ k }\]
Hint: shift the index and recall the Riemann Sum

Answer 2

Harmonic Series?

Answer 3

It certainly looks like one

Answer 4

If someone figures it out, can they please post the solution so that I can look at it later? I have to go to school right now.

Answer 5

\[\lim_{n \rightarrow \infty} \sum_{k=n+1}^{2n} \frac{ 1 }{ k }\]

using the hint, let \(k=n+t\),
then as \(k\to n+1\) we have \(t\to 1\), and
as \(k\to 2n\) we have \(t\to n\) :

\[\lim_{n \rightarrow \infty} \sum_{t=1}^{n} \frac{ 1 }{ n+t }\]

Answer 6

that sum an be rearranged as
\[\lim_{n \rightarrow \infty} \sum_{t=1}^{n} \frac{ 1 }{ 1+(t/n) } *\dfrac{1-0}{n}\]

Answer 7

can you eyeball the values of b, a and f(x) for the corresponding definite integral ?

Answer 8

would a=1, b= infinity, and f(x)= 1/(1+(1/n))?

Answer 9

Nope, try again.

Answer 10

are a and b right?

Answer 11

all wrong, i suggest you review riemann sum first...

Answer 12

This could help ( https://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#Rate_of_divergence):
\(\sum_{k=1}^{n} \frac1k = \log(n) + \gamma + \epsilon_n\).

Write \(\sum_{k=n+1}^{2n} \) as \(\sum_{k=1}^{2n} - \sum_{k=1}^n\).

Answer 13

sorry, I'm extremely confused as to what I'm supposed to do

Answer 14

I'm having a difficult time understanding the lower bound being in terms of t and the upper limit being n, and how that would turn into an integral.

Answer 15

\[\lim_{n \rightarrow \infty} \sum_{t=1}^{n} \frac{ 1 }{ 1+(t/n) } *\dfrac{1-0}{n} ~~=~~ \int\limits_0^1 \dfrac{1}{1+x}\,dx\]

Answer 16

to convince yourself more, maybe try expressing that integral as riemann sum again

Answer 17

Can you please explain to me how you got that integral?

Answer 18

have you reviewd riemann sum ?

Answer 19

wouldn't the riemann sum be 1/1 + 1/2 + 1/3 + 1/4 + ... ?

Answer 20

since t =1 and n is increasing by one integer each time?

Answer 21

the "\(x_i\)'s" are the \(t/n\), where \(n\) is fixed. -> 1/n, 2/n, ...,n/n.
The distance \(x_{i+1}-x_i = 1/n\).
notation as in https://en.wikipedia.org/wiki/Riemann_integral#Riemann_sums

Answer 22

There's nothing much to explain from @ganeshie8 's last 'mathematical answer', just review the Riemann integral.

Answer 23

ok, thanks

Answer 24

You might try Wolfram Alpha too..