# Finding $\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$

Find $$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$$

My work

$$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}=\frac{\sum_{m=1}^n\frac{\sqrt m}{m+1}}{\sqrt n}$$

The series $$\sum_{m=1}^\infty \frac{\sqrt m}{m+1}$$ does not converge so can I say $$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$$ does not exist?

• Where does the upper-limit $\infty$ in the sum come from? So you are saying $\lim_{n\to\infty}\frac{\overbrace{1+1+\dots+1}^{n}}{n}$ does not exist? Commented Jun 21, 2019 at 9:24
• Did you try Stolz Cesaro? Commented Jun 21, 2019 at 9:59

The series $$\sum_{m=1}^\infty \frac{\sqrt m}{m+1}$$ does not converge so can I say $$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$$ does not exist?

Because Zacky deleted his answer, I'll repeat the useful observation that divergence of a numerator does not mean the fraction necessarily diverges...

As for finding the limit; you can rewrite towards a Riemann sum: $$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}=\lim_{n\rightarrow\infty }\sum_{k=1}^{n}\frac{\sqrt k}{(k+1)\sqrt n}=\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\frac{k+1}{\sqrt {kn}}} \tag{\star}$$ Now you have an upper bound: $$(\star) : \lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\frac{k+1}{\sqrt {kn}}}\color{blue}{\le}\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{{\sqrt {\frac{k}{n}}}} = \int_0^1 \frac{1}{\sqrt{x}}\,\mbox{d}x = \color{blue}{2}$$ but also a lower bound: \begin{align}(\star) : \lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\frac{k+1}{\sqrt {kn}}} =\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt\frac{k^2+2k+1}{{kn}}} & \color{red}{\ge}\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{{\sqrt {\frac{k+3}{n}}}}\\ & =\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{m=4}^{n+3}\frac{1}{{\sqrt {\frac{m}{n}}}}\\[5pt] & = \int_0^1 \frac{1}{\sqrt{x}}\,\mbox{d}x = \color{red}{2}\end{align} So we have: $$\boxed{\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n} = 2}$$

• Thank you for your answer Commented Jun 21, 2019 at 10:11

$$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}=\lim_{n\rightarrow\infty }\sum_{r=1}^{n}\frac{\sqrt r}{(r+1)\sqrt n}$$

$$=\lim_{n\rightarrow\infty }\sum_{r=1}^{n}\frac{\sqrt{\frac rn}}{n\frac{(r+1)}{n}}$$ $$=\lim_{n\rightarrow\infty }\frac 1n\sum_{r=1}^{n}\frac{\sqrt{\frac rn}}{\frac rn + \frac1n}$$

$$=\int\limits_0^1\frac{\sqrt x}{dx+x}dx \approx \int\limits_0^1\frac{\sqrt x}{x}dx$$ $$=\boxed{2}$$

• Am I missing something or $\frac{r+1}{n} \ne \frac{r}{n}+1$...? Commented Jun 21, 2019 at 9:35
• I think that you have a mistake in the last expression before the integral. The denominator should be $\frac r n+\frac 1n$ and not $\frac r n+1$. Commented Jun 21, 2019 at 9:37
• edited...i fixed that Commented Jun 21, 2019 at 10:21
• So $\lim_{n\to\infty}\frac{1}{n}=dx$...? You get the right answer, but it looks more like physics than mathematics :-). Commented Jun 21, 2019 at 12:43
• Ah, I see... 😂 Commented Jun 21, 2019 at 13:16

By the theorem of Cesàro-Stolz (a discrete version of the l'Hopital rule for $$\frac\infty\infty$$), the quotient $$\frac{\sum_{m=1}^n a_m}{\sum_{m=1}^n b_m}$$ has a limit if the denominator grows to infinity and the quotient $$\frac{a_n}{b_n}$$ of the last terms has a limit, and then both limits have the same value.

Here $$\frac{a_n}{b_n}=\frac{\frac{\sqrt{n}}{n+1}}{\sqrt{n}-\sqrt{n-1}}=\frac{n+\sqrt{n(n-1)}}{n+1}\xrightarrow{n\to \infty} 2.$$

• Thank you so much. I did not know the theorem and it's a faster way to find this limit. Commented Jun 21, 2019 at 10:42

The fact that $$\sum_{m=1}^\infty \frac{\sqrt m}{m+1}$$ diverges does not mean $$\sum_{m=1}^n \frac{\sqrt m}{m+1}$$ does too.

For instance, the harmonic sum $$\sum_{m=1}^\infty \frac{1}{m}$$ is divergent, but the sum up to $$n$$, i.e., $$\sum_{m=1}^n \frac{1}{m}$$ is a rational number.