Limit in which terms involve n

Question

Limit n goes to infinity $\frac{1}{n}+\frac{1}{n+1}+......+\frac{1}{n+2n}$.

Well the answer came out log[3].. but i dont know how.? I am trying for MCA.

please don't tell complete answer just a hint would be useful

Just to clarify, you do you mean $$\lim_{n\to\infty}\sum_{i=0}^{2n}\frac{1}{n+i}$$ — Stella Biderman, Mar 27 '14 at 19:23
Okay, but the rest is right? Interesting, I've never seen this problem before. I was just asking because I found this a little hard to read. — Stella Biderman, Mar 27 '14 at 19:26

score 5 · Answer 1 · answered Mar 27 '14 at 19:27

5

Hint:

Note that

$$\sum_{i = 0}^{2n} \frac{1}{n + i} = \sum_{i = 0}^{2n} \frac 1 n \cdot \frac 1 {1 + i/n} = \sum_{i =0}^{2n} \frac 1 n f\left(\frac i n\right)$$

for $f(x) = \frac 1 {1 + x}$. Do you recognize a Riemann sum?

answered Mar 27 '14 at 19:27

ok as definite integral as sum of limits thanxx – Amit Mar 27 '14 at 19:28

score 0 · Answer 2 · answered Mar 27 '14 at 20:38

Draw a picture of $y=1/x$. Use the picture to note that $$\int_k^{k+1} \frac{dx}{x} \lt \frac{1}{k}\lt \int_{k-1}^k \frac{dx}{x}.$$ Thus $$\int_n^{3n}\frac{dx}{x}\lt \sum_n^{3n}\frac{1}{k}\lt \int_{n-1}^{3n-1} \frac{dx}{x}.$$ Evaluate the integrals, and Squeeze.

Limit in which terms involve n

2 Answers2