Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$i a_i = \left(i + 1\right) a_{i-1} + 2(i - 1)$ where $a_0 = a_1 = 0$

Solving this recurence relation.

How can I do this ? I tried to make something like

$i (a_i + 2(i-1)) = (i+1) (a_{i-1} + 2 (i-1))$ but find it kind of useless because i would actually change so I can't use a geometric sequence.

Does anyone know how to solve this recurrence relation ? Please show all of your steps. Thanks

share|cite|improve this question
up vote 1 down vote accepted

Notice that if you divide $ia_i=(i+1)a_{i-1}+2(i-1)$ by $i(i+1)$, you get


in which the two terms containing terms of the sequence have the same form. Let $b_i=\dfrac{a_i}{i+1}$; then $b_{i-1}=\dfrac{a_{i-1}}i$, and after you divide through by $i(i+1)$ your recurrence becomes


where of course I assume that $i>0$. This means that the $b_i$’s are just a sum of fractions of the form $\dfrac{2(i-1)}{i(i+1)}$ for consecutive values of $i$. And $b_0=0$, so

$$\begin{align*} b_n&=b_{n-1}+\frac{2(n-1)}{n(n+1)}\\ &=b_{n-2}+\frac{2(n-2)}{(n-1)n}+\frac{2(n-1)}{n(n+1)}\\ &=b_{n-3}+\frac{2(n-3)}{(n-2)(n-1)}+\frac{2(n-2)}{(n-1)n}+\frac{2(n-1)}{n(n+1)}\\ &\;\vdots\\ &=\sum_{k=1}^n\frac{2(k-1)}{k(k+1)}\\ &=2\sum_{k=1}^n\frac{k-1}{k(k+1)}\\ &=2\left(\sum_{k=1}^n\frac1{k+1}-\sum_{k=1}^n\frac1{k(k+1)}\right)\\ &=2\left(H_{n+1}-1-\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)\right)\\ &=2\left(H_{n+1}-1-\left(1-\frac1{n+1}\right)\right)\\ &=2\left(H_{n+1}+\frac1{n+1}-2\right)\;, \end{align*}$$

where $$H_n=\sum_{k=1}^n\frac1k$$ is the $n$-th harmonic number. Now recall that $a_n=(n+1)b_n$ to get the solution to the original recurrence:

$$\begin{align*} a_n&=(n+1)b_n\\ &=2(n+1)\left(H_{n+1}+\frac1{n+1}-2\right)\\ &=2\Big((n+1)H_{n+1}+1-2(n+1)\Big)\\ &=2\Big((n+1)H_{n+1}-2n-1\Big)\;. \end{align*}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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