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 $T_n=1+2+\cdots +n=\frac{n(n+1)}{2}.$
...
- 165k
2
votes
Accepted
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}\...
- 2,096
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
$$ p_n\left(1+\frac{\log(n+1)}{n}\right)=\left(1+\frac{\log(n+1)}{n}\right)^n-(n+1)-\frac{\log(n+1)}{n} < -\frac{\log(n+1)}{n} $$
...
- 340k
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 ...
- 5,971
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 $n$
$$H_n^{\left(\frac{1}{2}\right)}=2\sqrt{n}+\zeta \left(\frac{1}{2}\...
- 216k
1
vote
Convergence of $\sum_{n=1}^\infty \frac{\sum_{i=1}^n\frac{1}{ \sqrt i}}{n^2}$
Just for fun.
By the Hermite-Hadamard inequality
$$ \sqrt{4n+2}-\sqrt{6}=\int_{3/2}^{n+1/2}\frac{dx}{\sqrt{x}}\geq \sum_{k=2}^{n}\frac{1}{\sqrt{k}} $$
and by the inequalities fulfilled by the central ...
- 340k
1
vote
Convergence of $\sum \frac{b_n}{n}$ Where $b_n = 1, -1, -1, 1, 1, 1, -1, -1,-1,-1,1,1,1,1,1,....$
Let $S_k=b_1+b_2+\ldots +b_k.$
Denote $$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 ...
- 6,102
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