how can I proof this summation diverge

Question

$$\lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+.....+\frac{1}{n^2})$$ I tried to seperate it , and I could manage to proof this part does converge: $$\lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+.....+\frac{1}{2n})$$ But struggled to show divergence of the rest of the series by comparing it to the Harmonic series or using any of the divergence tests we`ve learned.

Thanks!

Answer 1

In that first sum, for a fixed $n$, take each of the first $n$ terms and compare to $\frac1{2n}$, take the next $n$ terms and compare to $\frac1{3n}$, and so on. Using this, what can you say that the entire sum is greater than?

Now let $n\to\infty$.

Answer 2

Since $\gamma:=\lim_{n\to\infty}(H_n-\ln n)$ is a finite value called the Euler-Mascheroni constant,$$\begin{align}\lim_{n\to\infty}(H_n-\ln n-\gamma)&=\lim_{n\to\infty}(H_{n^2}-2\ln n-\gamma)\\&=0\\\implies\lim_{n\to\infty}(H_{n^2}-H_n-\ln n)&=0\\\implies\lim_{n\to\infty}(H_{n^2}-H_n)&=\infty.\end{align}$$

Answer 3

$$\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right)+\cdots\left(\frac1{2^{m-1}+1}+\cdots \frac1{2^{m}}\right)+\cdots \\<\frac12+\frac12+\frac12+\cdots\frac12+\cdots$$

There are $m$ groups so that the sum exceeds $\dfrac m2$. Now if you add from $\dfrac1{n+1}=\dfrac1{2^m+1}$ to $\dfrac1{n^2}=\dfrac1{2^{2m}}$, the sum will exceed $\dfrac m2=\log_2\sqrt{n}$.

Answer 4

We can perform a crude bound by considering that for $x \ge 1$, we have $$0 < \frac{1}{\lceil x \rceil} \le \frac{1}{x} \le \frac{1}{\lfloor x \rfloor},$$ hence for integers $n \ge 1$ $$\sum_{k=2}^{n+1} \frac{1}{k} = \int_{x=1}^n \frac{1}{\lceil x \rceil} \, dx \le \int_{x=1}^n \frac{1}{x} \, dx = \log n \le \int_{x=1}^n \frac{1}{\lfloor x \rfloor} \, dx = \sum_{k=1}^n \frac{1}{k}.$$ If we denote $$H_n = \sum_{k=1}^n \frac{1}{k},$$ the above may be written $$H_{n+1} - 1 \le \log n \le H_n. \tag 1$$ Taking the LHS inequality, we have $$H_n \le 1 + \log(n-1), \quad n \ge 2$$ where all we have done is added $1$ to both sides and replaced $n$ with $n-1$ (thus requiring the condition $n \ge 2$ since the original inequality applies for $n \ge 1$). Therefore, we now have the bounds $$\log n \le H_n \le 1 + \log (n-1). \tag 2$$ Since your given sum is $$S(n) = \sum_{k={n+1}}^{n^2} \frac{1}{k} = H_{n^2} - H_n,$$ this yields the lower bound $$\log n^2 - (1 + \log(n-1)) \le S(n) .$$ This is because the lower bound of the difference is obtained by taking the lower bound of $H_{n^2}$ and subtracting the upper bound of $H_n$. Now we simplify and relax the lower bound further, giving $$S(n) \ge -1 + \log \frac{n^2}{n-1} > -1 + \log \frac{n^2}{n} = -1 + \log n,$$ consequently $S(n) \to \infty$ as $n \to \infty$. Our argument has been completely elementary, relying on integral calculus and the monotonicity of the logarithm, without needing to prove that $\lim_{n \to \infty} H_n - \log n$ exists.