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I got a question which says

$ 1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$

I got the solution by dividing by $7$ and subtracting it from original sum. Repeated for two times.(Suggest me if any other better way of doing this).

However now i am interested in understanding the series 1,2,4,7,11,..... In which the difference of the numbers are consecutive natural numbers.

How to find the sum of

$1+2+4+7+11+\cdots nterms$

This is my first question in MSE. If there are some guidelines i need to follow, which i am not, please let me know.

  • 0
    Thanks for the helpful link.2012-07-17

6 Answers 6

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Your series (without the denominators) is $\sum_{i=1}^n 1+ \frac {i(i-1)}2=\sum_{i=1}^n 1+\frac {i^2}2 - \frac i2=n+\frac {n(n+1)(2n+1)}{12}-\frac{n(n+1)}4$ This is an application of Faulhaber's formula

  • 0
    i/2 needs to be subtracted not added2012-07-17
6

Let's first find the general form of the terms of the sum $1 + 2+ 4 + 7 + 11 +\ldots.$ The terms obey the recurrence $a_n = a_{n-1} + n$, where $a_0 = 1.$ Using standard techniques we find $a_n = \frac{1}{2}n(n+1) + 1.$ These are basically the triangular numbers, as indicated by @anon in the comments.

The sum of the first $n+1$ terms can be found using Faulhaber's formula, as indicated by @RossMillikan, $\begin{eqnarray*} \sum_{k=0}^n a_k &=& \frac{1}{2}\sum_{k=0}^n n^2 + \frac{1}{2}\sum_{k=0}^n n + \sum_{k=0}^n 1 \\ &=& \frac{1}{2}\frac{1}{6}n(n+1)(2n+1) + \frac{1}{2}\frac{1}{2} n(n+1) + n+1 \\ &=& \frac{1}{6}(n+1)(n^2+2n+6). \end{eqnarray*}$

The sum $\begin{equation*} \sum_{n=0}^\infty \frac{a_n}{7^n} = \frac{1}{2}\sum_{n=0}^\infty \frac{n^2}{7^n} + \frac{1}{2}\sum_{n=0}^\infty \frac{n}{7^n} + \sum_{n=0}^\infty \frac{1}{7^n} \tag{1} \end{equation*}$ can be found by a standard trick. Consider the geometric series $\sum_{k=0}^\infty x^n = \frac{1}{1-x}$ for $|x|<1$. The last sum on the right hand side of (1) is just $\displaystyle\frac{1}{1-\frac{1}{7}} = 7/6$. Now notice that for $m\in \mathbb{N}$ $\begin{eqnarray*} \sum_{n=0}^\infty n^m x^n &=& \left(x \frac{d}{dx}\right)^m \sum_{k=0}^\infty x^n \\ &=& \left(x \frac{d}{dx}\right)^m \frac{1}{1-x}. \end{eqnarray*}$ Therefore, $\begin{eqnarray*} \sum_{n=0}^\infty \frac{a_n}{7^n} &=& \left[\frac{1}{2} \left(x \frac{d}{dx}\right)^2 \frac{1}{1-x} + \frac{1}{2} \left(x \frac{d}{dx}\right) \frac{1}{1-x} + \frac{1}{1-x} \right]_{x=1/7} \\ &=& \frac{1}{2}\frac{7}{27} + \frac{1}{2} \frac{7}{36} + \frac{7}{6} \\ &=& \frac{301}{216}. \end{eqnarray*}$

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    @Sandy: The trick you use to sum $1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$ is pretty good too!2012-07-18
3

Hint: your series is simply $\sum\limits_n^{k} \frac{n^2}{2}-\frac{n}{2}+1$

  • 0
    What is with the infinity at the top limit? What is the argument of the sum? This is unclear.2012-07-17
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Note that starting with your first element, $a_0=1$; to get to $a_1=2$ we sum $1$, to get to $a_2=4$, we sum $2$, to get to $a_3=7$, we sum $3$. So in general, we can say that

$a_{n}=a_{n-1}+n$

This is

$ \begin{align} 2&=1+1 \\[8pt] 4& =2+2\\[8pt] 7&=4+3\\[8pt] 11& =7+4\\[8pt] \cdots&=\cdots\\[8pt] a_n&=a_{n-1}+n \end{align} $

We can get several solutions to you problem, I will share $2$:

Solution 1 Use the recursion to obtain a closed formula:

Since we know that $a_{n}=a_{n-1}+n$ we can write

$a_{n-1}=a_{n-2}+n-1$

so we get $a_{n}=a_{n-2}+(n-1)+n$

Repeating this process, we get that

$a_{n}=a_{n-3}+(n-2)+(n-1)+n$

$a_{n}=a_{n-4}+(n-3)+(n-2)+(n-1)+n$

$a_{n}=a_{n-5}+(n-4)+(n-3)+(n-2)+(n-1)+n$

...so in general we can say that

$a_n=a_{n-k}+(n-k+1)+\cdots+n$ for any $k$ a natural number.

(actually we should be proving the above by induction, but let it be)

So we can choose $k=n$, which means...

$a_n=a_{n-n}+(n-n+1)+\cdots+n$

$a_n=a_0+(1+\cdots+n)$

$a_n=1+\frac{n(n+1)}{2}$

Solution 2

Starting from $a_{n}=a_{n-1}+n$ we use generating functions:

$\eqalign{ & \sum\limits_{n = 1}^\infty {{a_n}} {x^n} = \sum\limits_{n = 1}^\infty {{a_{n - 1}}} {x^n} + \sum\limits_{n = 1}^\infty n {x^n} \cr & \sum\limits_{n = 1}^\infty {{a_n}} {x^n} = x\sum\limits_{n = 1}^\infty {{a_{n - 1}}} {x^{n - 1}} + x\sum\limits_{n = 1}^\infty {n{x^{n - 1}}} \cr & \sum\limits_{n = 0}^\infty {{a_n}} {x^n} - {a_0} = x\sum\limits_{n = 0}^\infty {{a_n}} {x^n} + x\frac{d}{{dx}}\sum\limits_{n = 0}^\infty {{x^n}} \cr & A\left( x \right) - 1 = xA\left( x \right) + x\frac{d}{{dx}}\frac{1}{{1 - x}} \cr & A\left( x \right) - xA\left( x \right) = 1 + \frac{x}{{{{\left( {1 - x} \right)}^2}}} \cr & \left( {1 - x} \right)A\left( x \right) = 1 + \frac{x}{{{{\left( {1 - x} \right)}^2}}} \cr & A\left( x \right) = \frac{1}{{1 - x}} + \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr} $

The generating sequence of $f(x)=\frac{1}{{1 - x}}$ is $a_n=1$, while the generating sequence of $g(x)=\frac{x}{{{{\left( {1 - x} \right)}^3}}}$ can be obtained by diferentiation of the first one:

$\eqalign{ & \sum\limits_{n = 0}^\infty {{x^n}} = \frac{1}{{1 - x}} \cr & \sum\limits_{n = 1}^\infty {n{x^{n - 1}}} = \frac{1}{{{{\left( {1 - x} \right)}^2}}} \cr & \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){x^{n - 2}}} = \frac{2}{{{{\left( {1 - x} \right)}^3}}} \cr & \sum\limits_{n = 2}^\infty {\frac{{n\left( {n - 1} \right)}}{2}{x^{n - 1}}} = \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr & \sum\limits_{n = 1}^\infty {\frac{{n\left( {n + 1} \right)}}{2}{x^n}} = \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr} $

so we finally get that

$a_n=1+{\frac{{n\left( {n + 1} \right)}}{2}}$

ADD I forgot to add the sum of the $a_n$s!

You need to evaluate

$\sum_{k=0}^{n-1} a_k=\sum_{k=0}^{n-1} 1+\sum_{k=0}^{n-1}\frac{k(k+1)}{2}$

$\sum_{k=0}^{n-1} a_k=n+\sum_{k=0}^{n}\frac{k(k-1)}{2}$ $\sum_{k=0}^{n-1} a_k=n+\sum_{k=0}^{n}{k\choose 2}$

Using the binomial identity

$\sum_{k=0}^n {k\choose l}={{n+1}\choose {l+1}}$ we get

(you can find how to obtain it here

$\sum_{k=0}^{n-1} a_k=n+{n+1\choose 3}$

$\sum_{k=0}^{n-1} a_k=n+\frac{(n+1)n(n-1)}{6}$

which is what you wanted.

  • 0
    +$1$ - I'm surprised $n$obody else me$n$tioned the binomial form, which is the canonical way (at least for a programmer) of handling explicit values for 'simplical' numbers.2012-07-31
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Let T = 1 + 2/7 + 4/7^2 + 7/7^3 + ... ----> (1)

Then T/7 = 1/7 + 2/7^2 + 4/7^3 + ... ----> (2)

Now, (2)-(1) => 6T/7 = 1 + 1/7 + 2/7^2 + 3/7^3 ----> (3)

Hence, (1/7)(6T/7) = 1/7 + 1/7^2 + 2/7^3 ----> (4)

So, (3)-(4)=> (1 - 1/7)(6T/7) = 1 + 0 + 1/7^2 + 1/7^3 + ...

       => (6/7)(6T/7) = 1 + (1/7^2 + 1/7^3 + .... infinity)         => 36T/49 = 1 + (1/7^2)/(1 - 1/7)    [Formula = a/(1-r)]         => 36T/49 = 1 + (1/49)/(6/7)         => T = (49/36)(1 + (1/49)(7/6))         => T = 49/36 + (49/36)(1/49)(7/6)         => T = 49/36 + 7/216         => T = (294 + 7)/216         => T = 301/216  
  • 0
    Sorry, Din't read that thing!2012-07-31
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(Edit: Upps, I see now this is essentially solution (2) of Peter Tamaroff's answer, but because it's much shorter I just leave it here)

Your sequence can be separated into 2 sequences, where we add each pair:
$\begin{eqnarray} &1&2&4&7&11&16&\cdots & = a_k\\ \hline =&1&1&1&1&1&1&\cdots \\ +&0&1&3&6&10&15&\cdots \\ \hline \end{eqnarray}$

Then the partial sums are, beginning the index k at 1:
$\begin{eqnarray} &1&3&7&14&25&41&\cdots &=&s_k\\ \hline =&1&2&3&4&5&6&\cdots &= &&=&k\\ +&0&1&4&10&20&35&\cdots &=&\binom{1+k}{3}&=&{(k+1)!\over 3! (k-2)!}\\ \hline =&1&3&7&14&25&41&\cdots &=&s_k&=& k+ {(k+1)!\over 3! (k-2)!}\\ \end{eqnarray}$

The last formula can be simplified to $ s_k = k+ {(k+1)k(k-1) \over 6} = {6k+k^3-k\over 6} = k \cdot {k^2+5\over 6}$