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7 votes

For a positive integer $N \geq 2$, let

$$A_N := \Sigma_{n=2}^N \frac{1}{n};$$

$$B_N := \int\limits_{x=1}^N \frac{1}{x} dx$$

Which of the following statements is true?

- As $N \rightarrow \infty, \: A_N$ increases to infinity but $B_N$ coverages to a finite number
- $A_N < B_N$ and the difference decreases as $N \rightarrow \infty $
- $A_N < B_N < A_N +1$
- $B_N < A_N < B_N +1$
- As $N \rightarrow \infty, \: B_N$ increases to infinity but $A_N$ coverages to a finite number

6 votes

$$\begin{align*} & \text{We know , } \left ( \ln n \right ) < H_n < \left ( \ln n +1 \right ) \\ & \Rightarrow B_n < H_n < \left ( B_n + 1 \right ) \\ &\text{Given, } A_n = H_n - 1 \\ & \Rightarrow B_n < \left ( A_n +1 \right ) < \left ( B_n + 1 \right ) \\ \end{align*}$$

Option C

*For **more :** concrete mathematics by knuth.chapter $6$*

"As N keeps on increasing we see at N tending to infinity, we find that both the integral and the harmonic series sum evalautes to infinity as well..But it is a general matter of fact that continuous sum which is integral is obviously larger than discrete sum ..Here continuous sum is Bn and discrete sum is An."

- I am not saying this is wrong.

proving above statements would be mathematically acceptable.

3

4 votes

Let H_{N }= $\sum_{n=1}^{\infty } \frac{1}{n} =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ . . .$ is a popular HP which is divergent infinite series (divergent means limit of this function won't be a finite number).

It's divergent nature can be prove by 'Comparison Test' , which proves that sum of elements of this infinite series would be infinite.

https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Another way to prove it's divergent nature is by 'Integral Test' , in which we'll prove that $\int_{1}^{\infty } \frac {1}{x}dx$ would be logn and we know when n $\rightarrow \infty then $ logn $\rightarrow \infty$

a and e ) Both function A_{N }and B_{N} would tend to $\infty$ when N$\rightarrow \infty$

c and d ) By using map we can compare both function H_{N }and B_{N}

We can easily see if we find area using $\sum$ the rectangular will leads to some extra area than the area which will be covered using $\int_{}^{}$ so we can say from here H_{N} > B_{N}

in our question here A_{N} is discrete sum of our HP series , excluding 1 so its basically H_{N}-1

so we have proved yet that logn=B_{N} <H_{N}=A_{N} +1

Another way to prove lower bound and upper bound of H_{N }is like this

As $y=\frac {1}{x}$ is a Monotonically Decreasing function

means if some X ∈ [K,K+1] then $\frac{1}{k+1} \leq \frac{1}{x} \leq \frac{1}{K}$

So $\sum_{K=1}^{n}\frac {1}{K+1} \leq \int_{1}^{n}\frac{1}{x} \leq \sum_{K=1}^{n} \frac {1}{K}$

So A_{N} $\leq$ logn $\leq$ H_{N}

so eventually logn $\leq$H_{N} $\leq$logn+1

**Useful reference**

https://math.dartmouth.edu/archive/m105f13/public_html/m105f13notes1.pdf