search
Log In
0 votes
156 views

The series $\sum_{k=2}^{\infty} \frac{1}{k(k-1)}$ converges to

  1. $-1$
  2. $1$
  3. $0$
  4. does not converge
in Quantitative Aptitude
recategorized by
156 views

1 Answer

2 votes

Answer:(B)

$\displaystyle\sum _{k=2}^{\infty}\frac{1}{k(k-1)}$

$\displaystyle=\sum _{k=2}^{\infty}\frac{k-(k-1)}{k(k-1)}$

$\displaystyle=\sum _{k=2}^{\infty}\left(\frac{1}{(k-1)}-\frac{1}{k}\right)$

$=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots$

$=1-\frac{1}{\infty}$

$=1$

Related questions

0 votes
0 answers
1
125 views
Let $\{a_n\}$ be a sequence of non-negative real numbers such that the series $\Sigma_{n=1}^{\infty} a_n$ is convergent. If $p$ is a real number such that the series $\Sigma \frac{\sqrt{a_n}}{n^p}$ diverges, then $p$ must be strictly less than $\frac{1}{2}$ $p$ must be ... $1$ but can be greater than$\frac{1}{2}$ $p$ must be strictly less than $1$ but can be greater than or equal to $\frac{1}{2}$
asked Sep 23, 2019 in Others Arjun 125 views
0 votes
0 answers
2
76 views
The infinite series $\Sigma_{n=1}^{\infty} \frac{a^n \log n}{n^2}$ converges if and only if $a \in [-1, 1)$ $a \in (-1, 1]$ $a \in [-1, 1]$ $a \in (-\infty, \infty)$
asked Sep 13, 2018 in Others jothee 76 views
0 votes
1 answer
3
352 views
If $a,b$ are positive real variables whose sum is a constant $\lambda$, then the minimum value of $\sqrt{(1+1/a)(1+1/b)}$ is $\lambda \: – 1/\lambda$ $\lambda + 2/\lambda$ $\lambda+1/\lambda$ None of the above
asked Sep 23, 2019 in Quantitative Aptitude Arjun 352 views
1 vote
3 answers
4
432 views
Let $x$ be a positive real number. Then $x^2+\pi ^2 + x^{2 \pi} > x \pi+ (\pi + x) x^{\pi}$ $x^{\pi}+\pi^x > x^{2 \pi} + \pi ^{2x}$ $\pi x +(\pi+x)x^{\pi} > x^2+\pi ^2 + x^{2 \pi}$ none of the above
asked Sep 23, 2019 in Quantitative Aptitude Arjun 432 views
...