What should I do next?   I've tried to show that  $n\sum_{i=1}^n a_i &gt;= \sum_{i=1}^n a_i + n$ but it gets me to sth like:  $\sum_{i=1}^n a_i &gt;= \frac{n}{(n-1)}$  $a_1, a_2... \in R$
39105
How can I show that $n\sum_{i=1}^n {x_i^2} \ge (\sum_{i=1}^n{x_i})^2$ for any natural number $n$ and $x_i \in\mathbb{R}?$ I assume there is something about Cauchy-Schwarz and induction, but I really don't see it.
40528
$\sum_{i=1}^n(a_i)\sum_{i=1}^n(\frac{1}{a_i})\ge n^2$ is what I want to prove, where $a_i&gt;0$ and I need to showwhen equality holds.  I have tried using Cauchy-Schwarz, where inside the sums I multiply each value by 1 to do it but it doesn't work out.  I have also done  $\sum_{i=1}^n(a_i)\sum_{i=1}^n(\frac{1}{a_i})=\sum_{i=1}^n \sum_{i=1}^n(\frac{a_i}{a_i})=\sum_{i=1}^n\sum_{i=1}^n(1)=\sum_{i=1}^n(n)=n^2$, but this proves an equality.  Am I not allowed to do the first step I did in my work?
118919
If $a_n &gt;0$ for all $n\geq1$, show that the series  $$\sum_{n=1}^{\infty} \frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)}$$  converges.  Can someone check if my solution is correct:  $$\sum_{n=1}^{M}\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-\frac{1}{(1+a_1)(1+a_2)...(1+a_M)} &lt; 1 $$  Since the partial sum of the series is bounded, the series is convergent.
34773
I have to find the formula for sum $$\sum_{i=2}^{n} \frac1{i^2-1}$$ I remember reading somewhere that $\displaystyle \frac1{i^2-1}$ can be shown as $\displaystyle\frac1{i+1}$ and $\displaystyle \frac1{i-1}$.  If I can express it as this, then I should be able to create a telescopic sum, and produce the formula easily.  But how can I get to those two fractions in the first place so that I can build the proof?
101345
I tried getting it into a closed form but failed. Could someone help me out?  $$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$
30701
$$\sum_{i=1}^n \frac1{n+i} = \sum_{i=1}^{2n}\frac{(-1)^{i+1}}i,\quad \text{ for }n\ge 1.$$  I really have no clue how on to solve this one. It's the last one on my assignment. Would really enjoy some help with this one.
55436
$$\sum_{n=1}^{\infty} \frac{n}{3\cdot 5\cdot 7 \dots (2n+1)}$$  Can someone please help me in solving this problem, I tried to take the summation of the numerator and denominator individually but my teacher said that it is wrong to do the summation individually, can somebody please explain why is it wrong to take individual summation and please recommend the correct way of solving this problem
104205
I'm trying to find the sum of :   $$\sum_{i=1}^n\frac i {2^i}$$  I've tried to run $i$ from $1$ to $∞$ , and found that the sum is $2$ , i.e :   $$\sum_{i=1}^\infty\frac i {2^i} =2$$  since :  $$(1/2 + 1/4 + 1/8 + \cdots) + (1/4 + 1/8 + 1/16 +\cdots) + (1/8 + 1/16 + \cdots) +\cdots = 1+ 1/2 + 1/4 + 1/8 + \cdots = 2 $$  But when I run $i$ to $n$ , it's a little bit different , can anyone please explain ?   Regards
73778
I want to prove that if $\sum_{n=1}^\infty a_{2n}$ and $\sum_{n=1}^\infty a_{2n-1}$ converge then $\sum_{n=1}^\infty a_n$ converges as well.  Can I prove this like this: $\sum_{n=1}^\infty a_{2n-1}+\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_{2n-1}+a_{2n}=\sum_{n=1}^\infty a_n$  The first equality can be done because they both converge, but I'm not sure about the second one.
98650
Let $a_n$ be a sequence of complex numbers such that the sum from $n=1$ to infinity of $a_n/k^n$ is zero, for all k=1,2,3,... That is $$\sum_{n=1}^{\infty}\frac{a_n}{k^n}=0, k=1,2,3,...$$  Prove that $a_n=0$ for all n.  I want to use the identity theorem I think (it seems like it might work here), but I can't see how to apply it in this situation. Could anyone help with this please?
68440
$\displaystyle\sum_{i=0}^n2^i=2^{n+1}-1$.  I don't understand induction so I could use some help.
58285
If $a_1=1$ and $a_n=a_{n-1}+\dfrac{1}{a_{n-1}}$ for $n≥2$ , then prove that $12 &lt; a_{75} &lt; 15$ ?  I have tried solving this by:  $$a_{75} - a_1 = \frac{1}{a_1}+\frac{1}{a_2}....+\frac{1}{a_{74}}$$
34046
I was looking at: $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$  It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation?
120733
More specifically, I'm supposed to compute $\displaystyle\sum_{k=1}^{n} \frac 1 {k(k + 1)} $ by using the equality $\frac 1 {k(k + 1)} = \frac 1 k - \frac 1 {k + 1}$ and the problem before which just says that, $\displaystyle\sum_{j=1}^{n} a_j - a_{j - 1} = a_n - a_0$.  I can add up the sum for any $n$ but I'm not sure what they mean by "compute".   Thanks!
98452
Let $x_{1},x_{2},\cdots,x_{n}$ be real numbers. Show that     $$\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|.$$   I think this problem may be solved using nice methods, but I can't find them yet; I know this may be of use: $$|a|+|b|\ge |a+b|.$$ But I can't make it work. Thank you everyone.
116756
I don't do math a long time, so I completely don't remember how to prove that: $$ \sum_{i=1}^\infty \frac{i}{2^i} = 2 $$  Can anybody help me?
51742
I need to find the values of $a$ and $n$ in the following  $$ (1 + ax)^n  = \sum_{i=0}^\infty \binom{2i}{i}x^i$$  How can I compare things and show that $a=-4$ and $n = -{1 \over 2}$. It's easy to expand $1 \over \sqrt{1 -4x}$ see that it's valid but I'm looking other way around.