# Sum of the series : $1 + 2+ 4 + 7 + 11 +\cdots$

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.

• Do you mean the sum of the first $n$ terms of $1+2+4+7+11+\cdots$? Also, what is the rule for the terms. You say "In which the difference of the numbers are continuous natural numbers" but this makes no sense to me. – Alex Becker Jul 17 '12 at 1:50
• As an infinite series it is divergent, but there are a few ways to regularize it. @AlexBecker Look at the consecutive differences; the terms are the triangular numbers shifted up one. (This might explain the remark you quote too). – anon Jul 17 '12 at 1:51
• @AlexBecker difference of the numbers(2-1(1),4-2(2),7-4(3),11-7(4),...) Thats exactly what i mentioned.. Yes i am looking for the solution for this number series.. – mk.. Jul 17 '12 at 1:52
• @anon Thanks, I hadn't noticed that. Happy, again I'm unsure what you mean. You say "Yes i am looking for the solution for this number series" but series don't have solutions, they have sums, which can be either partial sums (the first $n$ terms) or the full sum, and in this case the full sum does not exist. – Alex Becker Jul 17 '12 at 1:54
• Will correct that. I want to find sum to nterms. – mk.. Jul 17 '12 at 1:56

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

• Correct me if i am wrong, the sum is not matching with the sum of series values. Ex: For n=1, the sum of the series you have given is 2 but the first term is 1. Pls clarify – mk.. Jul 17 '12 at 2:04
• It should be $i(i-1)/2$ on the LHS, and therefore a minus sign on the far rightmost term, to match the OP's terms – anon Jul 17 '12 at 2:08
• Thanks Ross, @anon. Changed the answer. Thank you – mk.. Jul 17 '12 at 2:24
• i/2 needs to be subtracted not added – Jorge Fernández Hidalgo Jul 17 '12 at 2:26

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, $$\begin{eqnarray*} \sum_{n=0}^\infty \frac{a_n}{7^n} &=& \left[\frac{1}{2} \left(x \frac{d}{dx}\right)^2 \frac{1}{1-x} + \frac{1}{2} \left(x \frac{d}{dx}\right) \frac{1}{1-x} + \frac{1}{1-x} \right]_{x=1/7} \\ &=& \frac{1}{2}\frac{7}{27} + \frac{1}{2} \frac{7}{36} + \frac{7}{6} \\ &=& \frac{301}{216}. \end{eqnarray*}$$

• @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! – user26872 Jul 18 '12 at 7:15

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.

• +1 - I'm surprised nobody else mentioned the binomial form, which is the canonical way (at least for a programmer) of handling explicit values for 'simplical' numbers. – Steven Stadnicki Jul 31 '12 at 17:18

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

• Thanks. Yup it was explained, it is one more than triangular numbers, and thats how we arrived at it. Thanks for posting it. – mk.. Jul 17 '12 at 2:26
• a very easy way to find it is by deriving it twice – Jorge Fernández Hidalgo Jul 17 '12 at 2:36
• What is with the infinity at the top limit? What is the argument of the sum? This is unclear. – Pedro Tamaroff Jul 17 '12 at 17:00

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

• Regarding this problem, the question states: "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)." This answer appears to suggest the same method. – Jonas Meyer Jul 31 '12 at 16:03
• Sorry, Din't read that thing! – Ashwyn Jul 31 '12 at 16:24

(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 $$s_k = k+ {(k+1)k(k-1) \over 6} = {6k+k^3-k\over 6} = k \cdot {k^2+5\over 6}$$