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$$ \lim_{x \to +\infty} \frac{1}{\sqrt{(x^2)}} + \frac{1}{\sqrt{(x^2+1)}} + \frac{1}{\sqrt{(x^2+2)}}......+ \frac{1}{\sqrt{(x^2+2x)}}$$ This is the given problem now I tried approaching it with definite integrals as sum of limit so $$ \lim_{x \to +\infty} \frac{1}{x.\sqrt{(1+\frac{r}{x^2})}}$$ but to make it integral r and x should be of same power so possibly this method would fail so how to proceed this problem?

Tesla
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2 Answers2

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Try putting

$$\sqrt{1+\frac n{x^2}}=\sqrt{1+\left(\frac{\sqrt n}{x}\right)^2}$$

Edit: What about trying a way that actually works? Let us apply the squeeze theorem:

$$2\xleftarrow[n\to\infty]{}\frac{2+\frac1n}{\sqrt{\frac2n+1}}=\frac{2n+1}{\sqrt{2n+n^2}}\le\sum_{k=0}^{2n}\frac1{\sqrt{k+n^2}}\le\frac{2n+1}{\sqrt{n^2}}=\frac{2n+1}n\xrightarrow[n\to\infty]{} 2$$

Timbuc
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  • But then how will be substitute integral variable ? r/n is substituted as variable , how will $\sqrt{r}/n $ be substituted – Tesla Oct 07 '14 at 15:02
  • You're right, this doesn't help, though the algebra is correct. I got confused by your use of symbols: $;x;$ is usually the variable's name whereas $;n,m,k;$ are usually reserved for the running index (over the naturals). – Timbuc Oct 07 '14 at 15:06
  • @Tesla, check the edit I've just made. – Timbuc Oct 07 '14 at 15:19
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Note that $x$ must be an integer. The terms in each sum $S_x$ are a decreasing family hence $$\frac{2x+1}{\sqrt{x^2+2x}}\leqslant S_x\leqslant\frac{2x+1}{\sqrt{x^2}},$$ which is enough to deduce that $\lim\limits_{x\to+\infty}S_x=2$.

Did
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