Answer by Witold for is the approach shown below in $\lim_{n \to \infty}...
Correct approach: $$n\frac{1}{\sqrt{n^2+n}} < \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}<n\frac{1}{\sqrt{n^2+1}}.$$Since there are limits $$\lim \limits_{n \to...
View ArticleAnswer by tbrugere for is the approach shown below in $\lim_{n \to \infty}...
Ok the thing you should understand to solve this is the result basically comes from the idea that $$\frac{1}{\sqrt{n^2 + i}}$$ is close to $$\frac{1}{\sqrt{n^2}}$$when $i \leq n$ and $n$ is big (and n...
View ArticleAnswer by CHAMSI for is the approach shown below in $\lim_{n \to \infty}...
Let $ n $ be a positive integer, we have the following :\begin{aligned} 1-\sum_{k=1}^{n}{\frac{1}{\sqrt{n^{2}+k}}}&=\sum_{k=1}^{n}{\left(\frac{1}{n}-\frac{1}{\sqrt{n^{2}+k}}\right)}\\...
View ArticleAnswer by MafPrivate for is the approach shown below in $\lim_{n \to \infty}...
Let the sum be $S$.$$\dfrac{n}{\sqrt{n^2+n}}< S < \dfrac{n}{\sqrt{n^2+1}} \\ \lim_{n\to\infty}\dfrac{n}{\sqrt{n^2+n}}\le \lim_{n\to\infty}S \le \lim_{n\to\infty}\dfrac{n}{\sqrt{n^2+1}} \\ 1 \le...
View Articleis the approach shown below in $\lim_{n \to \infty}...
The approach in a reference textbook:Let $f_{n}=\frac{n}{\sqrt{n^{2}+n}}$ Then $\lim_{n \to \infty } f_{n} = \frac{1}{\sqrt{1+\frac{1}{n}}}=1$From Cauchy's first theorem on limits:\begin{align*}\lim_{n...
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