$i,j\in \mathbb{N}$ and $i&lt;j $. Prove that $$\sum_{n=i}^{j}\frac{1}{n}$$ isn't integer using without using the .
70463
Assume that $k,n\in\mathbb{Z}^+$. Prove that the sum \begin{equation*} \dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+\ldots +\dfrac{1}{k+n-1}+\dfrac{1}{k+n} \end{equation*} is not an integer.   The case $k=0$ is proved in this question: ""
66163
I ran across this problem in some old notes, and I frustratingly can't figure out how to do it   Let $a_i$ and $b_i$ be sequences of natural numbers, use induction to show  $\sum_{i=1}^n (a_ib_i)^{1/2} \le (\sum_{i=1}^n a_i)^{1/2}(\sum_{i=1}^n b_i)^{1/2} $  Obviously this is trivial to show for n=1. I can't make much progress on n+1. I've tried various tactics, squaring both sides etc.   Any hint or help would be appreciated. Thanks.
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.
69440
Show by induction that for all natural numbers n>3  $(1+\frac{1}{n})^n&lt;n$  Let $(1+\frac{1}{n})^n&lt;n$ be true !   We show that $(1+\frac{1}{n+1})^{n+1})&lt;n+1$  $(1+\frac{1}{n})^n(1+\frac{1}{n})&lt;n(1+\frac{1}{n})$   $(1+\frac{1}{n})^{n+1}&lt;n+1$  So now we just need to show that   $(1+\frac{1}{n+1})^{n+1}&lt;(1+\frac{1}{n})^{n+1}$  So here , If I show that :   $(1+\frac{1}{n+1})&lt;(1+\frac{1}{n})$   Does that mean im done with exercise ?
31037
An  which I am not able to solve.  Let $\mathsf{S_{n}}$ denote the $n$-th partial sum of the series: $$\sum\limits_{r=1}^{\infty} \frac{1}{r(r+1)}$$  Prove that for every integer $k&gt;1$, there exists integers $m$ and $n$, such that $$s_{m}-s_{n} =\frac{1}{k}$$
120942
This is also a question on my exam paper that i proved by using mathematical induction. However, my tutor tells me that it can be proved without using mathematical induction. I really want to know how to deal with in another way.  Show that for any positive integer n, $(3n)!/(3!)^n$ is an integer.
14700
I oversaw a high-school mathematics test the other day, and one of the problems was the following     Show, using induction or other means, that $$\sum_{i = 1}^n\frac1{i(i+1)} = 1-\frac1{n+1}$$   The induction proof is very standard, where the induction step relies on the fact that $\frac{1}{n+1} + \frac{1}{n(n+1)} = \frac{1}{n}$, and I'm sure it's been answered on this site before. However, I got intrigued by the "or other means" part of the question.  I don't know whether the teacher who wrote the test even considered any alternative solutions (he may just have written it so that if anyone has a crazy idea that works out, then they should get a full score for it), but I tried to find one. For instance, we may do the following telescope-ish argument: $$ \frac{1}{1(1+1)} + \frac{1}{2(2+1)} + \cdots + \frac{1}{(n-1)n} + \frac{1}{n(n+1)} + \frac{1}{n+1}\\ = \frac{1}{1(1+1)} + \frac{1}{2(2+1)} + \cdots + \frac{1}{(n-1)n} + \frac1n\\ \vdots\\ = \frac{1}{1(1+1)} + \frac12\\ = 1 $$ However, I feel that this is just an induction proof in disguise (hidden in the vertical dots). If one uses the mechanics of the induction proof to check whether the formula is true for a specific $n$, then one certainly does the exact same calculations as I have done here.  Is there a proof of this fact that clearly does not use induction (or at least hides it better)? The more elementary the better, and the ultimate goal would be to do it within the syllabus of the students taking the test (or at least not far from it). For reference, they should be familiar with the summation formula of arithmetic and geometric series and their derivations (so techniques resembling those would be well within specifications).  If there is a solution using calculus, then the students should be able to integrate elementary trigonometric functions, as well as exponential functions, logarithms and rational functions. They are familiar with integration by parts, substitution and partial fractions.  I welcome more advanced solutions as well, of course.
22499
I need to show if $a$ is in $\mathbb{R}$ but not equal to $0$, and $a+\dfrac{1}{a}$ is integer, $a^t+\dfrac{1}{a^t}$ is also an integer for all $t\in\mathbb N$. Can you provide me some hints please?
53324
How can I prove that $$\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1 \,\,\, ?$$  I do know a way to prove this (see my answer) but I'm curious to know what other approaches could be taken in dealing with it.
75725
Prove that $$\sum_{i=1}^{\infty} \frac{1}{i(2i+3)} = \frac89 -\frac23\ln2$$  I tried using integration but failed miserably. Hints please.
26124
Can someone look at my proof. I am supposed to prove by induction. The question is to prove the following:  $$\sum _{i=0}^{n}{i} =\frac { n\left( n+1 \right)  }{ 2 } .$$  If $n=1$  Then  $$1=\sum_{i=1}^1{i}=\frac{1(1+1)}{2}=1$$  Now assume $n=k$.  Thus  $$\sum_{i=1}^k{i}=\frac{k(k+1)}{2}=1$$  If now $n=k+1$  $$\sum_{i=1}^{k+1}{i}=\sum_{i=1}^k{i}+(k+1)= \frac{(k+1)(k+2)}{2}$$  Is there anything more that I have to do, or is this it?
80310
How can I prove $$\sum_{i=1}^{n}(-1)^{i-1}\dbinom{n}{i}\frac{1}{i}=\frac{1}{1}+\frac{1}{2}+....+\frac{1}{n}\quad ?$$ I have tried induction, but it doesn't work.
68698
Do you have any idea to prove the following identity via a combinatorial (or algebraic) method?  $\sum_{n=r}^{n=\infty} {n \choose r}^{-1} = \frac{r}{r-1}$  This is  of the book Chen C.C., Koh K.M. Principles and techniques in combinatorics. The book does not give a solution, although it mentions: "see H. W. Gould, Combinatorial Identities, Morgantown, W.V. (1972), 18-19".  many thanks in advance, Shahram
118912
Let $p,q,r$ be integers such that the symmetric sum of fractions $\dfrac{pq}{r}+\dfrac{qr}{p}+\dfrac{pr}{q}$ is an integer. Prove that each of the numbers: $\dfrac{pq}{r},\dfrac{qr}{p},\dfrac{pr}{q}$ is an integer. How to do this? Someone posted but deleted soon.
39355
$T(0)=1 \\ T(n) = 1 + \sum_{j=0}^{n-1}T(j) \\ $  Show that $T(n) = 2^n$.  I know how to prove this by induction, but I would like to know how to show this using first principles.  Edit: The way I want to solve this problem is manipulate $T(n)$ in such a way that it ends up as $2^n$.
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!
2269
Give a combinatorial proof for this identity for non-negative integer $k$ and $n$ such that $0 \leq k &lt; n$   $$\sum_{j=0}^{k} \binom{n}{j} = \sum_{j=0}^k \binom{n-1-j}{k-j}2^j$$   My attempt: I tried to reduce this identity. Since $k,n$ are nonnegative integer and $k&lt;n$ then I can change the upper limit into  $$\sum_{j=0}^{n-1} \binom{n}{j} =2^n-1 \quad \text{and} \quad \sum_{j=0}^{n-1} \binom{n-1-j}{n-1-j}2^j = \sum_{j=0}^{n-1} 2^j ~.$$  So, I just need to show $$\sum_{j=0}^{n-1} 2^j =2^n-1 $$ using combinatorial proof. Am I right? I really need helps