7
$\begingroup$

Is there anything known about the value of the series $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\frac{1}{1+2+3+4+5+6}+\cdots$ ?

$\endgroup$

2 Answers 2

23
$\begingroup$

The denominators $a_n$ are $$ a_n = \sum_{k=1}^{n}k = \frac{n(n+1)}{2} $$ so $$ \sum_{n=1}^{\infty} \frac1{a_n} = \sum_{n=1}^{\infty} \frac{2}{n(n+1)} = 2\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 2 $$ If you don't have the resources to check this on a computer (out to infinity), you will have your computer or calculator carry out the calculation for $n<N$. You should then get a number near the partial sum $$ s_N=\sum_{n=1}^{N-1} \frac1{a_n} = 2 \left( 1 - \frac{1}{N} \right), $$ where whatever difference you find between the computer's answer and this expected answer will be the result of truncation or rounding errors due to the machine representation of each floating point number (which usually involve binary mantissas as in IEEE formats) encountered along the way.

$\endgroup$
1
  • $\begingroup$ whats your email address please, im not smart enough to figure it out from your profile $\endgroup$ Nov 21, 2012 at 22:59
18
$\begingroup$

Yes, in fact, everything is known about it. Step 1: get a formula for the denominators (I assume the 6th one is supposed to have a 6 in it, not stop at 5 like the 5th one). Step 2: apply partial fractions to get a telescoping series. Enjoy.

$\endgroup$
3
  • $\begingroup$ thanks! fixed the typo after the 5yh term. $\endgroup$
    – jimjim
    Feb 9, 2012 at 11:15
  • $\begingroup$ It equals 2.... $\endgroup$
    – Xonatron
    Feb 9, 2012 at 14:41
  • 1
    $\begingroup$ @Gerry I think the last word was meant to be "profit". $\endgroup$
    – Pedro
    Apr 18, 2012 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.