Found problems: 119
2019 LIMIT Category C, Problem 7
The value of
$$\left(1+\frac26+\frac{2\cdot5}{6\cdot12}+\frac{2\cdot5\cdot8}{6\cdot12\cdot18}+\ldots\right)^3$$
1980 IMO, 4
Prove that $\sum \frac{1}{i_1i_2 \ldots i_k} = n$ is taken over all non-empty subsets $\left\{i_1,i_2, \ldots, i_k\right\}$ of $\left\{1,2,\ldots,n\right\}$. (The $k$ is not fixed, so we are summing over all the $2^n-1$ possible nonempty subsets.)
2004 VJIMC, Problem 2
Evaluate the sum
$$\sum_{n=0}^\infty\operatorname{arctan}\left(\frac1{1+n+n^2}\right).$$
2019 Jozsef Wildt International Math Competition, W. 53
Compute $$\lim \limits_{n \to \infty}\frac{1}{n}\sum \limits_{k=1}^n\frac{\sqrt[n+k+1]{n+1}-\sqrt[n+k]{n}}{\sqrt[n+k]{n+1}-\sqrt[n+k]{n}}$$
2010 VJIMC, Problem 1
a) Is it true that for every bijection $f:\mathbb N\to\mathbb N$ the series
$$\sum_{n=1}^\infty\frac1{nf(n)}$$is convergent?
b) Prove that there exists a bijection $f:\mathbb N\to\mathbb N$ such that the series
$$\sum_{n=1}^\infty\frac1{n+f(n)}$$is convergent.
($\mathbb N$ is the set of all positive integers.)
1980 IMO Longlists, 16
Prove that $\sum \frac{1}{i_1i_2 \ldots i_k} = n$ is taken over all non-empty subsets $\left\{i_1,i_2, \ldots, i_k\right\}$ of $\left\{1,2,\ldots,n\right\}$. (The $k$ is not fixed, so we are summing over all the $2^n-1$ possible nonempty subsets.)
2019 Jozsef Wildt International Math Competition, W. 30
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[*] Prove that $$\lim \limits_{n \to \infty} \left(n+\frac{1}{4}-\zeta(3)-\zeta(5)-\cdots -\zeta(2n+1)\right)=0$$
[*] Calculate $$\sum \limits_{n=1}^{\infty} \left(n+\frac{1}{4}-\zeta(3)-\zeta(5)-\cdots -\zeta(2n+1)\right)$$
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2016 IMC, 3
Let $n$ be a positive integer. Also let $a_1, a_2, \dots, a_n$ and $b_1,b_2,\dots, b_n$ be real numbers such that $a_i+b_i>0$ for $i=1,2,\dots, n$. Prove that
$$\sum_{i=1}^n \frac{a_ib_i-b_i^2}{a_i+b_i}\le\frac{\displaystyle \sum_{i=1}^n a_i\cdot \sum_{i=1}^n b_i - \left( \sum_{i=1}^n b_i\right) ^2}{\displaystyle\sum_{i=1}^n (a_i+b_i)}$$.
(Proposed by Daniel Strzelecki, Nicolaus Copernicus University in Toruń, Poland)
2011 VTRMC, Problem 3
Find $\sum_{k=1}^\infty\frac{k^2-2}{(k+2)!}$.
2019 LIMIT Category C, Problem 3
Which of the following series are convergent?
$\textbf{(A)}~\sum_{n=1}^\infty\sqrt{\frac{2n^2+3}{5n^3+1}}$
$\textbf{(B)}~\sum_{n=1}^\infty\frac{(n+1)^n}{n^{n+3/2}}$
$\textbf{(C)}~\sum_{n=1}^\infty n^2x\left(1-x^2\right)^n$
$\textbf{(D)}~\text{None of the above}$
1995 IMC, 8
Let $(b_{n})_{n\in \mathbb{N}}$ be a sequence of positive real numbers such that $b_{0}=1$,
$b_{n}=2+\sqrt{b_{n-1}}-2\sqrt{1+\sqrt{b_{n-1}}}$. Calculate
$$\sum_{n=1}^{\infty}b_{n}2^{n}.$$
2015 BMT Spring, 7
Evaluate $\sum_{k=0}^{37}(-1)^k\binom{75}{2k}$.
1983 IMO Shortlist, 21
Find the greatest integer less than or equal to $\sum_{k=1}^{2^{1983}} k^{\frac{1}{1983} -1}.$
2006 IMO Shortlist, 2
The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by \[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]Show that $ a_{n} > 0$ for all $ n\geq 1$.
[i]Proposed by Mariusz Skalba, Poland[/i]
2016 Bundeswettbewerb Mathematik, 1
There are $\tfrac{n(n+1)}{2}$ distinct sums of two distinct numbers, if there are $n$ numbers.
For which $n \ (n \geq 3)$ do there exist $n$ distinct integers, such that those sums are $\tfrac{n(n-1)}{2}$ consecutive numbers?
1978 IMO, 2
Let $f$ be an injective function from ${1,2,3,\ldots}$ in itself. Prove that for any $n$ we have: $\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.$
2001 VJIMC, Problem 2
Prove that for any prime $p\ge5$, the number
$$\sum_{0<k<\frac{2p}3}\binom pk$$is divisible by $p^2$.
2016 IMC, 1
Let $(x_1,x_2,\ldots)$ be a sequence of positive real numbers satisfying ${\displaystyle \sum_{n=1}^{\infty}\frac{x_n}{2n-1}=1}$. Prove that $$ \displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{k} \frac{x_n}{k^2} \le2. $$
(Proposed by Gerhard J. Woeginger, The Netherlands)
1978 Putnam, B6
Let $p$ and $n$ be positive integers. Suppose that the numbers $c_{hk}$ ($h=1,2,\ldots,n$ ; $k=1,2,\ldots,ph$) satisfy $0 \leq c_{hk} \leq 1.$ Prove that
$$ \left( \sum \frac{ c_{hk} }{h} \right)^2 \leq 2p \sum c_{hk} ,$$
where each summation is over all admissible ordered pairs $(h,k).$