7 votes

Convergence of $\sum \frac{b_n}{n}$ Where $b_n = 1, -1, -1, 1, 1, 1, -1, -1,-1,-1,1,1,1,1,1,....$

This is equivalent to whether: $$\sum_{n=0}^\infty (-1)^n c_n$$ converges, where $$c_n=\sum_{k=T_{n}+1}^{T_{n+1}}\frac1{k}=\sum_{j=1}^{n+1}\frac{1}{T_n+j}$$ Where $T_n=1+2+\cdots +n=\frac{n(n+1)}{2}.$ ...
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2 votes
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Root in $(1,2]$ of Equation $x^n-x-n=0$

First, I shall prove a stronger claim. Claim: $u_n \geq 1 + \frac{2}{n}$ for $n \geq 2$. Proof: Applying IVT on $\left(1 + \frac{2}{n}, 2\right)$, we simply have to evaluate $f_n\left(1 + \frac{2}{n}\...
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2 votes

Root in $(1,2]$ of Equation $x^n-x-n=0$

An alternative approach: let $p_n(x)=x^n-x-n$. We may notice that $$ p_n\left(1+\frac{\log(n+1)}{n}\right)=\left(1+\frac{\log(n+1)}{n}\right)^n-(n+1)-\frac{\log(n+1)}{n} < -\frac{\log(n+1)}{n} $$ ...
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2 votes

Convergence of $\sum \frac{b_n}{n}$ Where $b_n = 1, -1, -1, 1, 1, 1, -1, -1,-1,-1,1,1,1,1,1,....$

The $n$-th "run" consists of $n$ elements (numerators of $1$ for odd $n$ and numerators of $-1$ for even $n$). The last element of the $n$-th run has index $n(n+1)/2$, and the first element ...
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1 vote

Convergence of $\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$

Using generalized harmonic numbers, you face $$S_p=\sum_{n=1}^p \frac{H_n^{\left(\frac{1}{2}\right)}}{n^2}$$ For large values of $n$ $$H_n^{\left(\frac{1}{2}\right)}=2\sqrt{n}+\zeta \left(\frac{1}{2}\...
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1 vote

Convergence of $\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$

Just for fun. By the Hermite-Hadamard inequality $$ \sqrt{4n+2}-\sqrt{6}=\int_{3/2}^{n+1/2}\frac{dx}{\sqrt{x}}\geq \sum_{k=2}^{n}\frac{1}{\sqrt{k}} $$ and by the inequalities fulfilled by the central ...
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1 vote

Convergence of $\sum \frac{b_n}{n}$ Where $b_n = 1, -1, -1, 1, 1, 1, -1, -1,-1,-1,1,1,1,1,1,....$

Let $S_k=b_1+b_2+\ldots +b_k.$ Denote $$u_n={2n(2n-1)\over 2}=n(2n-1)\quad v_n={(2n+1)2n\over 2}=(2n+1)n$$ Then $$S_{u_n}=n,\qquad S_{v_n}=-1$$ Moreover $S_k$ is decreasing for $u_n\le k\le v_n$ and ...
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