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Problem from the 1960 Putnam Olympiad involving the sum of a series
Assume that (1) holds, let us show that the range of sums is indeed an interval.
As you described in your part of the proof, if the range of all possible sums is an interval then it is an interval $[0 ...
- 925
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$A\subset\mathbb{N};A_N=\lvert A\cap\{1,\ldots,N\}\rvert.$ Does $\sum_{n\in A} \frac{1}{n}$ diverges $\implies\sum\frac{A_N}{N^2}$ diverges?
Everything is positive, so we can interchange order of summation as we like. Write $$ \sum_{N=1}^\infty \frac{A_N}{N^2} = \sum_{N=1}^\infty\ \sum_{n \in A \cap \{1,...,N\}}\ \frac{1}{N^2}.$$
The ...
- 7,651
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Find a closed formula for $\sum_{n=1}^\infty n^4{x^{n-1}}$
just follow Anne Bauval's comment:
$$\sum_{n=0}^{\infty} x^n = \frac1{1-x} \implies \sum_{n=0}^{\infty} nx^{n-1} = \frac1{(1-x)^2} \implies \sum_{n=0}^{\infty} nx^{n} = \frac x{(1-x)^2} \implies\\ \...
- 1,310
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Test the convergence of series ${1\over\sqrt n} - \frac{\sqrt n}{\sqrt{n+1}}$
$$\lim_{n\to\infty}\frac{1}{\sqrt n} - \frac{\sqrt n}{\sqrt{n+1}}=-1$$
The series diverges by the $n$th term test.
- 2,028
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Can I bound this somehow?
No. Consider the counterexample
$$x = 0.10110101101010101...$$
the concatenation of $2^k + 1$ ones alternating with $2^k$ zeros for all whole k.
For a length C, there are less than $\log_2{C}$ more ...
- 189
1
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Accepted
Question about the Thue-Morse Sequence and its relation with the Fabius function
Yes, because if we split the Thue-Morse sequence into pairs, they all will be either $01$ or $10$, so for any even $n$ we have $S(n)=0$, and for odd $n$ $S(n)$ differs by $1$.
WolframAlpha is right ...
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