I am to prove that if $$1-1/2+1/3-1/4+1/5-1/6+...=s$$ that $$1-1/2+1/3+1/5-1/4+1/7+1/9-1/6+...=3s/2$$    My approach:  $s/2=1/2-1/4+1/6-1/8+1/10-1/12...$ so that  $s+s/2=3s/2=1-(1/2-1/2)+1/3-(1/4+1/4)+1/5-(1/6-1/6)...=$ $1+1/3-1/2+1/5...$    This sort of leads to what I want to approach, but the terms are not quite in the right order. I assume I may not change the order of the last set of terms for it to match what I'm looking for. Any other suggestions?
122727
I need to show that $\sum (-1)^{n+1} / n$ = $k$ (which is $1 - 1/2 + 1/3 - 1/4+...$) can be expanded to a sum with $ 1 - 1/2 + 1/3 + 1/5 - 1/4 +1/7 + 1/9 - 1/6 + .... $ which is $3k/2$.   How can I start? I know that $k/2$ = $\sum (-1)^{n+1}/2n$ How can I go further so I get $3k/2$?
94783
How can I prove that $\sum^n_{i=1} \frac{1}{i(i+1)} = \frac{n}{n+1}$? I noticed that in the sum, the denominator has terms that cancel out, but I'm not sure how to take advantage of that.
83974
How do I simplify the following series $$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ \frac{1}{1+2+3+4} + \frac{1}{1+2+3+4+5} + \cdot\cdot\cdot + \frac{1}{1+2+3+\cdot\cdot\cdot+n}$$  From what I see, each term is the inverse of the sum of $n$ natural numbers. Assuming there are $N$ terms in the given series, $$a_N = \frac{1}{\sum\limits_{n = 1}^{N} n} = \frac{2}{n(n+1)}$$ $$\Rightarrow \sum\limits_{n=1}^{N} a_N = \sum\limits_{n=1}^{N} \frac{2}{n(n+1)}$$ ... and I'm stuck.  I've never actually done this kind of problem before (am new to sequences &amp; series).  So, a clear and detailed explanation of how to go about it would be most appreciated.  PS- I do not know how to do a telescoping series!!
58852
has the identity $$S(n+1, k+1) = \sum_i {n \choose i} S(i, k)\;.$$ I can't seem to find a proof for this and was wondering what it entails. Is it a purely combinatoric argument, or an algebraic manipulation of the explicit formula $S(n, k) = \frac{1}{k!} \sum_{i=0}^k (-1)^i {k \choose i}(k-i)^n $ with the known recurrence formula $S(n+1, k+1) = S(n, k) + (k+1)S(n, k+1)? $
49817
Lets define the following series $$S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.$$ I know that $S_0$ does not converge so let's suppose $n \in N$ and define $S_n$ for some $n$. We have $S_1=1$ , $S_2=\frac1 4$ , $S_3=\frac 1 {18}$ , $S_4=\frac 1{96}$ , $S_5=\frac 1{600}$ etc.. the numerator in all results is 1the pattern in denominator is $[1,4,1,96,600...]$ and   that is equal to $n*n!$ Finally I want to prove the general equality: $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$
105834
For a given prime $p$ and integer $k$,  $k∈&lt;1, p-1&gt;$ $S(k)$ is defined as a sum of remainders of $k, k^2, k^3, ..., k^{p-1}$ when divided by $p$. Prove that $S(k)=\frac{1}{2}p(p-1)$ is true for an odd number of different k. I managed to prove that $S(p-1)=\frac{1}{2}p(p-1)$ but can't see how to do it further.
111461
I need to use the identity $\frac{1}{k^2-1}=\frac{1}{2}(\frac{1}{k-1}-\frac{1}{k+1})$ to evaluate $\sum_{k=2}^n \frac{1}{k^2-1}$.  I am confused about how to begin this proof.  Also, how am I supposed to use the identity here?  Do I need to use induction?
82765
My problem is how to prove  $$\frac{1+2}{1}  &gt; \frac{1+3+\frac{3^2}{2!}}{1+2} &gt; \frac{1+4+\frac{4^2}{2!}+\frac{4^3}{3!}}{1+3+\frac{3^2}{2!}}&gt;\cdots $$  i am sure that the sequence is decreasing as limit exp(1).  However, the reason why the sequence is decreasing is not evident.  Thanks if you consider this issue and make me relief.
98404
Prove via induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$ Having a very difficult time with this proof, have done pages of work but I keep ending up with 1/(k+2).  Not sure when to apply the induction hypothesis and how to get the result $1- \frac{1}{(n+2)!}$.  Please help! thanks guys, youre the greatest!
87603
This is the Grandi's series: $1-1+1-1+1-1+\dots$  The series can be equal to $0$   $$(1-1)+(1-1)+(1-1)+\dots=0+0+0+\dots=0,$$  or to $1$   $$1-(1-1)-(1-1)-(1-1)-\dots=1-0-0-0\dots=1,$$  or to $1/2$   $$S=1-1+1-1+1-\dots,\quad\quad    S=1-(1-1+1-1+1-1+1-...)$$ $$\Rightarrow   S=1-S\Rightarrow     2S=1\Rightarrow    S=1/2$$  Isn't this a contradiction? The integers are closed under addition and subtraction, but we get a fraction. Why?
111071
I am having trouble with (a) and (b):  I evaluated (a) and I got that sun of the first three terms does not equal $\ln(5/8)$. I got that it is equal to $\ln(3/4) + \ln(2/3) + \ln(15/16)$ which is not equal to $\ln(5/8)$.  I am also having trouble with part (b). I cannot figure out how to calculate the math to cancel out the $(k+1)^2$ and turn it into $2(n  +1)$.  Any help would be amazing.
98791
$$ 2n^3 + 5 | 2n^4 +2n +2 - 2n^4 - 5n$$  $$= 2n^3+5 | 2-3n$$  $$3n-2≥2n^3 + 5$$  is this correct? Is there a more efficient way?
56919
prove if we want that the sum of some fractions be $1$ and the denominators of one of them is $d$ then another denominators should divisible by $d$ or $d$ should be divisible to another denominators.  It seems to be easy I tried to prove it.I first tried some cases.  $1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$  Here we can see that $6$ is divisible by $3$. Also here $6$ is divisible by $2$ .But I want to prove one of the denominators but here two of them is possible.After trying a lot I cannot found any proofs.Any hints?  update1: the numerator should be prime
471
How to prove for $s&lt;1$$$|a+b|^s\le|a|^s+|b|^s$$ I tried to prove $|a+b|\le (|a|^s+|b|^s)^{\frac{1}{s}}$, but $\frac{1}{s}$ may not be integer, so I do not know how to expand it if it is not integer.  Is it the right way to prove it by trying to expand it using binomial expansion? How to prove it?  Thanks!
99875
Let $a,b,c,d\in \mathbb{R}$ and $a+b+c+d=2$. Prove that $$\frac{a}{a^2-a+1}+\frac{b}{b^2-b+1}+\frac{c}{c^2-c+1}+\frac{d}{d^2-d+1}\le \frac{8}{3}.$$   We have $$\frac{a}{a^2-a+1}\le \frac{4}{3}a\Longleftrightarrow \frac{-a\left(2a-1\right)^2}{3\left(a^2-a+1\right)}\le 0,$$ (Right)  So $$LHS\le \frac{4}{3}\left(a+b+c+d\right)=\frac{8}{3}=RHS$$  When $a=b=c=d=\frac{1}{2}$  But also $a=b=c=1;d=-1$ i don't know how to solve it.
88116
Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}}  +\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} &gt; 24$$   My work:   Rationalizing the denominator gives   $$\frac{\sqrt{3}-1}{2}  +\frac{\sqrt{7}-\sqrt{5}}{2}+......+\frac{\sqrt{9999}-\sqrt{9997}}{2} .$$  Now by taking two as common and separating the positive and negative terms gives  $$\frac{1}{2} [ \{\sqrt{3} +\sqrt{7}+\dots +\sqrt{9999}\} - \{1+\sqrt{5} +\dots+\sqrt{9997}\}].$$  Can we do like this please suggest. Thanks.
69970
The main question is :  Prove that $n^n&gt;1\times3\times5\times7\times\dots\times(2n-1)$  My approach :  We can write the R.H.S as,  $$\frac{(2n-1)!}{2\times4\times6\times\dots\times2(n-1)}$$ We can write $(2n-1)!$ as $(2n-3)!(2n-1)2(n-1)$  Thus, in this manner we can cancel out even terms till $n$ appears. I'm having trouble doing this. What I intend to do is say that any positive number's factorial is less than the number raised to itself. I welcome any alternate method. Please help.  P.S. I am still in high school and do not understand binomial theorem and induction.