I need to prove that $b_n = \frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \dotsb + \frac{1}{\sqrt{n^2 + n}} \to 1$ as $n \to \infty$  My initial thoughts are to sandwich this (seems sensible to me). Finding the above inequality $b_n &lt; 1$ was easy, since clearly $\frac{1}{\sqrt{n^2 + k}} &lt; \frac{1}{n}$ for integer k. But I'm stuck on how to compare it with something from below.   A hint I've been given is that if $a_n \to a$, then $\sqrt{a_n} \to \sqrt{a}$ as $n \to \infty$, though I'm struggling to see a way of applying this.
121818
We have to show that  $$ \lim_{n\to\infty} \left( \dfrac 1 {\sqrt{n^2 +1}} +\dfrac 1 {\sqrt{n^2 +2}} + \dfrac 1 {\sqrt{n^2 +3}} + \ldots + \dfrac 1 {\sqrt{n^2 +n}} \right) = 1$$  Now, I thought of doing it as a definite integral, however the terms don't translate into a function of $\left(\dfrac r n\right)$. I tried searching wolfram-alpha, but it used Hurwitz-zeta function, of which I have no clue. (This is an assignment problem)
15491
Let $a,b,c$ are positive number such that $abc=1$. Prove that:   $$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\;\geq\; a+b+c$$   This problem froms my Math teacher. I have attempted to let $$(a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})$$. The inequality is equivalent to: $\frac{\sqrt{x^2-xy+y^2}}{y}+\frac{\sqrt{y^2-yz+z^2}}{z}+\frac{\sqrt{x^2-xz+z^2}}{x}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$  Then, I tried to use AM-GM but I stucked on it.
98003
Let   $$  a_n =  \left( 1 - \frac{1}{\sqrt 2} \right)  \dotsm  \left( 1 - \frac{1}{\sqrt {n+1}} \right),  \quad  n\geq 1. $$   Then $\lim_{n \to \infty} a_n$         is $1$,   is $0$,   does not exist,   is $\frac{1}{\sqrt\pi}$.      I tried this using the sandwich theorem as $$  0 \leq \sum_{r=1}^{n} \frac{-1}{\sqrt r+1} \leq \frac{-n}{\sqrt {n+2}} $$  which goes to zero. I am not sure though.
41812
Is there a way to show that $\sqrt{p_{n}} &lt; n$?  In , I show that $f_{2}(x)=\frac{x}{ln(x)} - \sqrt{x}$ is ascending, for $\forall x\geq e^{2}$. As a result, $\forall n \geq 3$ $$\frac{p_{n}}{ln(p_{n})} - \sqrt{p_{n}}\leq \frac{p_{n+1}}{ln(p_{n+1})} - \sqrt{p_{n+1}}$$ Also (and as a result), $\forall n \geq 3$ $$ \frac{p_{n}}{ln(p_{n})} - \sqrt{p_{n}} &gt; 0$$ Or $$ \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} &lt; \frac{\pi (p_{n})}{\sqrt{p_{n}}}$$  According to  $$\displaystyle\smash{\lim_{n \to \infty }}\frac{\pi (p_{n})}{p_{n}/ln(p_{n})}=1$$ Or, $\forall \varepsilon &gt;0$, $\exists N(\varepsilon )$: $\forall n&gt;N(\varepsilon )$ $$1- \varepsilon &lt; \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} &lt; 1+ \varepsilon$$ Or $$1- \varepsilon &lt; \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} &lt; \frac{\pi (p_{n})}{\sqrt{p_{n}}}$$ As a result $\forall \varepsilon &gt;0$, $\exists N(\varepsilon )$: $\forall n&gt;N(\varepsilon )$ $$(1 - \varepsilon ) \cdot  \sqrt{p_{n}} &lt; \pi (p_{n}) = n$$  But this is not enough.  Interestingly,  is true iff function $f_{4}(x)=\pi (x) - \sqrt{x}$ is strictly ascending ($x &lt; y \Rightarrow f(x) &lt; f(y)$) for prime arguments.  If $f_{4}(p_{n}) &lt; f_{4}(p_{n+1})$ then $$\pi (p_{n}) - \sqrt{p_{n}} &lt; \pi (p_{n+1}) - \sqrt{p_{n+1}}$$ Or $$\sqrt{p_{n+1}} - \sqrt{p_{n}} &lt; \pi (p_{n+1}) - \pi (p_{n}) =1$$  And vice-versa, if $$\sqrt{p_{n+1}} - \sqrt{p_{n}} &lt; 1$$ Then $$-\sqrt{p_{n}} &lt; -\sqrt{p_{n+1}} + 1$$ Or $$\pi (p_{n})-\sqrt{p_{n}} &lt; \pi (p_{n}) + 1 -\sqrt{p_{n+1}} = \pi (p_{n+1}) -\sqrt{p_{n+1}}$$  So, if Andrica's conjecture is true then $\forall n \geq 3$ $$\pi (p_{n})-\sqrt{p_{n}} &gt; 0$$ Or $$\sqrt{p_{n}} &lt; \pi (p_{n})= n$$
36572
I would like to show that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$ holds for all natural numbers. I got stuck here:  $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}.$  I would appreciate your help.
21748
I am having issues solving these problems: I tried using $(|a|+|b|)^2 \geq 0$ and $(|a|-|b|)^2 \geq 0$ but I am having problems constructing the proof. I need to gain some intuition on how to proceed.     Show that for $a,b\in (-\infty,+\infty)$,   the following inequalities apply:   a) $2|ab| \leq a^2 + b^2$  b) $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$
26008
Suppose $\sum{a_n}$ converges. How do I prove that $\sum{\frac{\sqrt{a_n}}{n}}$ converges.  So I know that just because  $\sum{a_n}$ converges, do not mean I can say anything about the converges of its square root. So I know that I can prove that if  $\sum{\sqrt{a_n}}$ converges then $\sum{\frac{\sqrt{a_n}}{n}}$ converges by Abel's Test, but I do not know where to start for the case where  $\sum{\frac{\sqrt{a_n}}{n}}$ diverges
42387
Kazakstan 2012 Suppose that $a, b \in\mathbb{R}$, and $a,b&gt;0$. If $\frac{1}{a}+\frac{1}{b}=2$ prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$ My idea : $a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqrt{ab}}{ab-a^2b^2}$, but where  can I use the fact that $\frac{1}{a}+\frac{1}{b}=2?$
117645
Prove that the inequality $$\left|\frac{m}{n}-\frac{1+\sqrt{5}}{2}\right|&lt;\frac{1}{mn}$$ holds for positive integers $m, n$ if and only if $m$ and $n$ where $m &gt; n$ are two successive terms of the Fibonacci sequence.   I thought about using the explicit formula for the Fibonacci sequence, which is $$F_n = \dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right),$$ but this seems to get computational. Is there an easier way to think about this?
13042
Trying to prove that above stated question for $n \geq 1$. A hint given is that you should use $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$. Using this, I think I reduced it to $\frac{1}{2}(\frac{1}{n^2}-(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}))$. Just not sure if it's correct, and what to do with the second half.
15333
How to prove that $ 2&lt;(1+\frac{1}{n})^{n}$ for every integer $n&gt;1$. I was thinking by induction, it works for $n=2$ but then I couldn't move forward.
24134
Let $a$ and $b$ be positive numbers such that $a+b=1$. Prove that: $$a^{4b^2}+b^{4a^2}\leq1$$ I think this inequality is very interesting because the equality &quot;occurs&quot; for $a=b=\frac{1}{2}$ and also for $a\rightarrow0$ and $b\rightarrow1$. I tried to work with a function of one variable, but the derivative is not easy. I also don't get something solvable by Taylor series.
45932
Let$a_n,b_n\in\mathbb R$ and $(a_n+b_n)b_n\neq 0\quad \forall n\in \mathbb{N}$. The series $\sum_{n=1}^\infty\frac{a_n}{b_n} $ and $\sum_{n=1}^\infty(\frac{a_n}{b_n})^2 $ are convergent.  How to prove that$$\sum_{n=1}^\infty\frac{a_n}{b_n+a_n} $$ is convergent.  Thanks in advance
48765
Having the following inequality  $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}&lt;2$$ To prove it for all natural numbers is it enough to show that:  $\frac{1}{(n+1)^2}-\frac{1}{n^2} &lt;2$  or  $\frac{1}{(n+1)^2}&lt;2-\frac{1}{n^2} $
77410
Apostol's Calculus, exercise number I 4.7 13.     Prove that if $n \geq 1$, then   $$ 2(\sqrt{n+1} - \sqrt{n}) &lt; \frac{1}{\sqrt{n}} &lt; 2(\sqrt{n} - \sqrt{n-1}) $$   and use this to prove that if $m \geq 2$, then   $$ 2 \sqrt{m} - 2 &lt; \sum_{n = 1}^m \frac{1}{\sqrt{n}} &lt; 2\sqrt{m} - 1 $$   In particular, when $m = 10^6$ the sum lies between $1998$ and $1999$.   After establishing the trivial case, for $n=1$, I can't think of a way to perform the inductive step:  assuming $2(\sqrt{n+1} - \sqrt{n}) &lt; \frac{1}{\sqrt{n}} &lt; 2(\sqrt{n} - \sqrt{n-1})$ and using this, deducing the same for $n+1$ is true as well:  $2(\sqrt{n+2} - \sqrt{n+1}) &lt; \frac{1}{\sqrt{n+1}} &lt; 2(\sqrt{n+1} - \sqrt{n})$  Any help is very appreciated, thanks
84235
I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:  \begin{align*} \lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&amp;=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\ &amp;=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4} \end{align*}  where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?
5998
Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} &gt; \frac{13}{24}$ for $n&gt;1,n\in\Bbb N$  To solve it I used induction but it is leading me nowhere my attempt was as follows:  Lets assume the inequality is true for $n = k$ then we need to prove that it is true for $k+1$  so we need to prove $\frac1{k+2} + \frac1{k+3}+\cdots+\frac1{2(k+1)} &gt; 13/24$  I don't know where to go from here please help.