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Calculus Level 5

Find the value of $S$ as defined below:

$S=1+\dfrac2{10}+\dfrac3{10^2}+\dfrac{10}{10^3}+\dfrac{22}{10^4}+\dfrac{51}{10^5}+\dfrac{125}{10^6}+\dfrac{293}{10^7}+ \cdots$

Enter 0.6666 if you come to the conclusion that the series fails to converge. Enter your Answer upto 4 decimal places

Note: The numerators follow the pattern $T_{n+3} = T_{n+2} + 2 T_{n+1} + 3 T_n$ .

The answer is 1.2428.

1 solution

Otto Bretscher
May 8, 2016

The numerators satisfy the recursion $a_{n+3}=a_{n+2}+2a_{n+1}+3a_{n}$ for $n>0$ . The generating function is found to be $\sum_{n=1}^{\infty}a_n x^n=\frac{1+x-x^2}{1-3x^3-2x^2-x}$ for $|x|<0.42$ . For $x=\frac{1}{10}$ the value is $\approx \boxed{1.2428}$

sir, Is it easy to figure out the recursion relation ?

Sabhrant Sachan - 5 years, 1 month ago

That was the harder part! Based on your last problem, I assumed (well, hoped) that you might have used a linear recursion involving the three preceding terms.

Another fun problem!

Otto Bretscher - 5 years, 1 month ago

The exact value is $1090/877$ .

D G - 3 years, 9 months ago

Keep 'em Coming

The answer is 1.2428.

1 solution

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