Simple High school Series & Limits

Calculus Level 4

$\large a_{n}=\sum_{k=1}^{n}\frac{k}{2^k}$ Suppose we define the sequence $a_n$ as above with natural number (n), find the value of

$\lim_{n\rightarrow \infty }\frac{2^n(6-3a_{n})}{\sqrt{n^2+5n+1}}$

1 solution

Vishwak Srinivasan
Jul 18, 2015

$a_n = \dfrac{1}{2^1} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \ldots + \dfrac{n}{2^n}$

$\dfrac{1}{2} a_n = 0 + \dfrac{1}{2^2} + \dfrac{2}{2^3} + \ldots + \dfrac{n-1}{2^n} + \dfrac{n}{2^{n+1}}$

Subtracting the two equations above, gives us:

$\Rightarrow \dfrac{1}{2}a_n = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \ldots + \dfrac{1}{2^n} - \dfrac{n}{2^{n+1}}$

$\Rightarrow \dfrac{1}{2}a_n = \dfrac{\frac{1}{2}\left(1-\frac{1}{2^n}\right)}{\frac{1}{2}} - \dfrac{n}{2^{n+1}}$

$\Rightarrow \dfrac{1}{2} a_n = 1 - \dfrac{1}{2^n} - \dfrac{n}{2^{n+1}}$

$\Rightarrow 3a_n = 6 - \dfrac{6}{2^n} - \dfrac{3n}{2^n}$

$\Rightarrow 2^n(6-3a_n) = 6 + 3n$

$\displaystyle \lim_{n\to \infty} \dfrac{2^n(6-3a_n)}{\sqrt{n^2 + 5n + 1}} = \lim_{n \to \infty} \dfrac{6+3n}{\sqrt{n^2 + 5n + 1}} = \boxed{3}$

Heads of you… I have tried a lot but not getting the answer but you have done some nice job…

Himanshu Agrawal - 5 years, 10 months ago

I love the way you converted the series into required form. Awesome .

Anurag Pandey - 4 years, 8 months ago