A SMALL FIBONACCI SUM

13 05 2010

If F_{n} is the nth Fibonacci Number then:

\displaystyle \sum_{i=0}^{n}{i \cdot F_{2i}} = n \cdot F_{2n+1} - F_{2n}

This identity can be easily proved using the method of induction with the basic recurrence relation of Fibonacci Numbers.

How can we find methods for constructing new identities like this one?


References:

[1]-Wikipedia – Fibonacci number
[2]-Chandra, Pravin and Weisstein, Eric W. “Fibonacci Number.” From MathWorld–A Wolfram Web Resource – http://mathworld.wolfram.com/FibonacciNumber.html

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TETRAHEDRAL NUMBERS RECIPROCALS SUM

25 12 2009

TETRAHEDRAL NUMBERS SERIES:

This post follows with the exercises on special numbers reciprocals related series, after the blog entries about Square Pyramidal Numbers and Polygonal Numbers . In fact, this example it is not very much interesting, but I wanted to write it before to deal with more difficult problems.

\displaystyle T_{n}=\frac{n(n+1)(n+2)}{6}=\binom{n+2}{3}

\displaystyle S(n)=\sum_{k=1}^{n}{\frac{1}{T_{k}}}=\sum_{k=1}^{n}{\frac{6}{k(k+1)(k+2)}}

If we split the main fraction into others:

\displaystyle \frac{S(n)}{6}=\sum_{k=1}^{n}{\frac{1}{k(k+1)(k+2)}}=\sum_{k=1}^{n}{\left( \frac{A}{k}+\frac{B}{k+1}+\frac{C}{k+2} \right) }

Solving the linear system of equations it gives:

\displaystyle A=\frac{1}{2} \; ; B=-1 \; ; C=\frac{1}{2};

This three series can be summed easily with the aid of the Harmonic Numbers:

\displaystyle \sum_{k=1}^{n}{\frac{1}{k}}=1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n}=H_n

\displaystyle \sum_{k=1}^{n}{\frac{1}{k+1}}=\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n}+\frac{1}{n+1}=H_n-1+\frac{1}{n+1}

\displaystyle \sum_{k=1}^{n}{\frac{1}{k+2}}=\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}=H_n-1-\frac{1}{2} + \frac{1}{n+1}+\frac{1}{n+2}

If we sustitute everything in the expression for the reciprocals sum:

\displaystyle \frac{S(n)}{6}=\frac{n}{n+1}-\frac{1}{2}-\frac{1}{4} +\frac{1}{2(n+1)}+\frac{1}{2(n+2)}

In the previous step we can see what does exactly means to be a “telescoping series“, the term H_n, has vanished and there is no need to handle the Euler Mascheroni Gamma and the Digamma Function:

H_{n}=\gamma + \psi_{0}(n+1)

Then the formula for the n-th partial sum is:

\displaystyle S(n)=\frac{3n(3+n)}{2(1+n)(2+n)}

And taking the limit we get:

\displaystyle S(\infty)=\lim_{n \leftarrow \infty}{S(n)}=\frac{3}{2}


References:[1]-Tetrahedral Number at- Wikipedia
[2]-Weisstein, Eric W. “Tetrahedral Number.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/TetrahedralNumber.html
[3] A000292-Tetrahedral (or pyramidal) numbers: C(n+2,3) = n(n+1)(n+2)/6. The On-Line Encyclopedia of Integer Sequences!






CURIOUS SERIES-001

16 08 2009

Curious Series 001

There´s a very common finite series, that use to be, at the begining on every book:

\displaystyle S_{n}(z)=\sum _{k=0}^n z^k =\frac{z^{n+1}-1}{z-1}

Where z can be real or complex.

There is a, very well known, particular case of this series where z=2:

\displaystyle S_{n-1}(2)=\sum _{k=0}^{n-1} 2^{k} =2^{n}-1=M_n

\displaystyle M_n are the Mersenne Numbers, and due to this sum, is easy to see that the Mersenne numbers consist of all 1s in base-2 (they are base 2 repunits)

But this entry is about another particular case of this finite sum:

\displaystyle S_{n}(\textbf{i})=\sum _{k=0}^{n} \textbf{i}^k

Where: \textbf{i}=\sqrt{-1}, is the complex unit.

\displaystyle S_{n}( \textbf{i} ) =\frac{1}{2}(1+\textbf{i}) \left(1-\textbf{i}^{n+1}\right)

This sum shows periodical behaviour with a period of 4, and its values changes from one vertex to another in a square of side equal to 1, if we plot them in the complex plane:

\displaystyle S_{n}( \textbf{i} )=\{1,1+\textbf{i},\textbf{i},0,1,1+\textbf{i},\textbf{i},...\}

\displaystyle S_{n}(\textbf{i})=1\; if \;n\equiv 0\;mod\;4

\displaystyle S_{n}(\textbf{i})=1+\textbf{i}\; if \;n\equiv 1\;mod\;4

\displaystyle S_{n}(\textbf{i})=\textbf{i}\; if \;n\equiv 2\;mod\;4

\displaystyle S_{n}(\textbf{i})=0\; if \;n\equiv 3\;mod\;4

If we take a look at the real part of the complex number S_{n}(\textbf{i}):

\displaystyle Re\bigg(\sum _{k=0}^{n} \textbf{i}^k\bigg)=\{1,1,0,0,1,1,0,0,...\}

Then we had just found the sequence A133872 from OEIS, and then we can construct another expressions for this sequence, and also for the problem series:

\displaystyle A133872(n)=Re\bigg(\sum _{k=0}^{n} \textbf{i}^k\bigg)

\displaystyle A133872(n)=\frac{1}{2}\bigg(\sum _{k=0}^{n} \textbf{i}^k + \sum _{k=0}^{n} \textbf{i}^{-k}\bigg)

\displaystyle A133872(n)=\frac{1}{2}+\frac{1}{2} \text{cos}\left(\frac{n \pi }{2}\right)+\frac{1}{2} \text{sin}\left(\frac{n \pi }{2}\right)

Then, if we expand to trigonometrical functions S_{n}( \textbf{i} ):

\displaystyle S_{n}( \textbf{i} ) =\left(\frac{1}{2}+\frac{1}{2} \text{cos}\left(\frac{n \pi }{2}\right)+\frac{1}{2} \text{sin}\left(\frac{n \pi }{2}\right)\right) + \textbf{i} \left( \frac{1}{2}-\frac{1}{2} \text{cos}\left(\frac{n \pi }{2}\right)+\frac{1}{2} \text{sin}\left(\frac{n \pi }{2}\right)\right)

And finally using the information inside OEIS:

\displaystyle S_{n}( \textbf{i} )= \text{mod}\left(\bigg\lfloor\frac{n+2}{2}\bigg\rfloor,2\right)+ \textbf{i}\cdot \text{mod}\left(\bigg\lfloor\frac{n+1}{2}\bigg\rfloor,2\right)\cdot \textbf{i}


References:[1]-A133872-Period 4: repeat 1,1,0,0. The On-Line Encyclopedia of Integer Sequences!






SQUARE PYRAMIDAL NUMBERS RECIPROCALS SUM

11 04 2009


This post continues with the work on some special, sets of integers, related series:

The square pyramidal numbers expression can be found on [1] and it is:

\displaystyle P(n)=\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}

The Square pyramidal numbers reciprocal finite serie is:

\displaystyle S(n)=\sum_{k=1}^n{\frac{6}{k(k+1)(2k+1)}}

Splitting the main fraction into others:

\displaystyle \frac{S(n)}{6}=\sum_{k=1}^n{\frac{1}{k(k+1)(2k+1)}}=\sum_{k=1}^n{\frac{A}{k}+\frac{B}{k+1}+\frac{C}{2k+1}}

Solving the equations gives:

\displaystyle A=1 ;\; B=1;\; C=-4;

Substituing the series (ii), and the expression for the Harmonic Numbers (i):

\displaystyle \frac{S(n)}{6}= \sum_{k=1}^n {\frac{1}{k}}+\sum_{k=1}^n{\frac{1}{k+1}}-4\sum_{k=1}^n{\frac{1}{2k+1}}=2H_n-\frac{n}{n+1}-4\sum_{k=1}^n{\frac{1}{2k+1}}

Taking into account the expressions: (i),(iv) and (vi), for the digamma function:

\displaystyle \frac{S(n)}{6}= 2\gamma+2\psi_{0}(n+1)-\frac{n}{n+1}-4\sum_{k=1}^n{\frac{1}{2k+1}}

\displaystyle S(n)=6[ 2\gamma+2\psi_{0}(n+1) -\frac{n}{n+1}-2[\psi_{0}(n+\frac{3}{2})+\gamma+2\log{2}-2]]

\displaystyle S(n)=6[2\psi_{0}(n+1)-2\psi_{0}(n+\frac{3}{2})-\frac{n}{n+1} -4\log{2}+4]

Calculating the limit, and using (vii):

\displaystyle S(\infty)= \lim_{x \to{+}\infty}{S(n)}= 6(3-4\log{2})=18-24\log{2}\approx 1.3644676665...

(see reference [4])


NOTES:

(i) Harmonic Numbers and Digamma Function at integer values:

\displaystyle H_{n}=\sum_{k=1}^n {\frac{1}{k}}=\gamma+\psi_{0}(n+1)

(ii) Changing series into Harmonic Numbers:

\displaystyle \sum_{k=1}^n{\frac{1}{k+1}} =\sum_{k=1}^n{\frac{1}{k}}-1+\frac{1}{n+1}=H_{n}-\frac{n}{n+1}

(iii) Changing one series into another:

\displaystyle \sum_{k=1}^n{\frac{1}{2k-1}} -\sum_{k=1}^n{\frac{1}{2k+1}}=1-\frac{1}{2n+1}

(iv) Digamma function at half-integer values:

\displaystyle \psi_{0}(n+\frac{1}{2})= -\gamma-2\log{2}+2\sum_{k=1}^n{\frac{1}{2k-1}}

(v) A Digamma function property:

\displaystyle \psi_{0}(z+1)= \psi_{0}(z)+\frac{1}{z}

(vi) Another expression for Digamma function at half-integer values:

Using (iii), (iv) and (v):

\displaystyle  \psi_{0}(n+\frac{3}{2}) = \psi_{0}(n+\frac{1}{2}) +\frac{1}{n+\frac{1}{2}}=\psi_{0}(n+\frac{1}{2})+\frac{2}{2n+1}= \\ = -\gamma-2\log{2} +2(1-\frac{1}{2n+1}+\sum_{k=1}^n{\frac{1}{2k+1}} )= \\= -\gamma-2\log{2}+2+2\sum_{k=1}^n{\frac{1}{2k+1}}

(vii) Limit for Digamma function at half-integer values:

\displaystyle L_{1}(\infty)=\lim_{x \to{+}\infty}{L_{1}(n)}= \lim_{x \to{+}\infty}{[\psi_{0}(n+1)-\psi_{0}(n+\frac{3}{2})] }

\displaystyle L_{1}(n)=\psi_{0}(n+1)-\psi_{0}(n+\frac{3}{2})=2\log{2} + \sum_{k=1}^n {\frac{1}{k}}-2 \sum_{k=1}^n {\frac{1}{2k-1}}

\displaystyle L_{1}(n)=2\log{2}-\sum_{k=1}^n {\frac{1}{k(2k-1)}}

The last serie limit can be derived from the Mercator-Mengoli infinite series for \displaystyle \log{2}\;. See [3].
This proof is interesting enough for another entry on the blog.

\displaystyle \log{2}=\sum_{k=1}^\infty {\frac{1}{2k(2k-1)}}

So:

\displaystyle L_{1}(\infty)=0


References:

[1]-Square Pyramidal Numbers @ Wikipedia Square Pyramidal Numbers
[2]-Weisstein, Eric W. “Digamma Function.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/DigammaFunction.html
[3]-Collection of formulae for log 2-Numbers Constants and Computation-Xavier Gourdon and Pascal Sebah.
[4]-A159354-Decimal expansion of 18-24*log(2) The On-Line Encyclopedia of Integer Sequences!
[5]-A000330-Square pyramidal numbers The On-Line Encyclopedia of Integer Sequences!


Archives:

[a]-041109-SQUARE PYRAMIDAL NUMBERS RECIPROCALS SUM.nb





INVERSE POLYGONAL NUMBERS SERIES-Notes

19 03 2009

The final result, in the preceeding post, can not be derived from a telescoping series [3], if \displaystyle k is not integer (See comments at reference [1]).

\displaystyle \sum_{n=1}^\infty \frac{1}{n(n+k)}=\frac{H_k}{k}

This lack of generality, can be avoided, if we consider a more general definition for the Harmonic Numbers [4], extended to the complex plane, using the function:

\displaystyle H_z=\gamma+\psi_0(z+1)

Where \displaystyle \psi_0 \; is the so called digamma function, and \displaystyle \;\gamma\; is the Euler-Mascheroni constant.

If you take a look at the expresion (15), in the reference [2] : We can find that one asymptotic expansion for the digamma function is:

\displaystyle \psi_0(k+1)=-\gamma+\sum_{n=1}^\infty{\frac{k}{n(n+k)}}

This is why the Polygonal Numbers Series sum is working:

\displaystyle H_k=\gamma-\gamma+ \sum_{n=1}^\infty{\frac{k}{n(n+k)}}

\displaystyle \frac{H_k}{k}=\sum_{n=1}^\infty{\frac{1}{n(n+k)}}=\frac{\gamma+\psi_0(k+1)}{k}

And the polygonal numbers infinite sum, can be expressed (if \displaystyle \;s\neq4\;) as:

\displaystyle S_{\infty}(s)=\frac{2}{4-s}*(\gamma+\psi_{0}\left(\frac{2}{s-2}\right))

This expresion works for all \displaystyle s>2, as well as for all nonreal \displaystyle s, It also works for all \displaystyle s<2, except if \displaystyle s<2, and \displaystyle s is \displaystyle \;\;0, 1, 4/3, 6/4, 8/5, 10/6, ... \;, because \displaystyle \;\psi_0\; is not defined for negative integers (See reference) [1]


References:

[1]-Charles R Greathouse IV – Comments @ My Math Forum Inverse Polygonal Series
[2]-Weisstein, Eric W. “Digamma Function.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/DigammaFunction.html
[3]-Telescoping Series @ Wikipedia Telescoping Series
[4]-Sondow, Jonathan and Weisstein, Eric W. “Harmonic Number.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/HarmonicNumber.html
[5]-Photo Martin Gardner, Mathematical Games, Scientific American, 211(5):126-133, taken from http://bit-player.org/2007/hung-over





INVERSE POLYGONAL NUMBERS SERIES

9 03 2009

Let there be \displaystyle P(n,s) the n-th Polygonal Number [4] of s sides, then:

\displaystyle P(n,s)=\frac{ (s-2)*n^{2}-(s-4)*n }{2}

For s=3, we get the formula for Triangular Numbers

\displaystyle P(n,3)=\frac{n^{2}+n}{2}=T_n

and with s=4 then we get the Squares:

\displaystyle P(n,4)=n^{2}

Polygonal numbers hold for the next identity:

\displaystyle P(n,s+1)=P(n,s)+ P(n-1,3)=P(n,s)+ T_{n-1}

The sum of the inverse of the first k polygonal numbers with side s, is:

\displaystyle S_{k}(s)=\sum_{n=1}^k{} \frac{1}{P(n,s)}

And its infinite series sum:
\displaystyle S_{\infty}(s)=\sum_{n=1}^{\infty{}} \frac{1}{P(n,s)}

Than this infinite series is convergent, can be proved, using some convergence test:

Raabe´s convergence Test:

\displaystyle \rho=\displaystyle\lim_{n \to{+}\infty}{n*\left(\frac{P(n+1,s)}{P(n,s)}-1\right) }=2>1

A series with a lower number of sides, upper bounds, series with higher sides: So the convergence in triangular numbers, implies the convergence of the remaining polygonal numbers series:

If \displaystyle s_1>s_2 \longrightarrow{} \frac{1}{P(n,s_1)}<\frac{1}{P(n,s_2)} \longrightarrow{} S_{\infty}(s_1)

The inverse series with Triangular numbers is a telescoping series [3]:

\displaystyle S_{\infty}(3)=2\cdot\sum_{n=1}^{\infty{}} \frac{1}{n\cdot(n+1)}= 2

The squares sum is the so called Basel Problem, [1] [2], related with the Riemann Zeta Function:

\displaystyle S_{\infty}(4)=\sum_{n=1}^{\infty{}} \frac{1}{n^2}=\zeta(2)=\frac{\pi^2}{6}

If \displaystyle s\neq4, then:

The series is a more general case of a telescoping series [3], related with the Harmonic Numbers.

\displaystyle S_{\infty}(s)=2\cdot \sum_{n=1}^{\infty{}} \frac{1}{(s-2)\cdot n^{2}-(s-4)*n}=\frac{2}{s-2}\cdot\sum_{n=1}^{\infty{}}\frac{1}{n\cdot(n+\frac{4-s}{s-2})}

\displaystyle \sum_{n=1}^{\infty{}} \frac{1}{n*(n+k)}=\frac{H_k}{k}

\displaystyle S_{\infty}(s)= \frac{2}{4-s}\cdot H_{\frac{4-s}{s-2}}

Example: In the hexagonal numbers case:

\displaystyle S_{\infty}(6)= -H_{-\frac{1}{2}} \approx 1.38629


Archives:

[a]-031009-INVERSE POLYGONAL NUMBERS SERIES.nb


References:

[1]-Estimating Basel Problem@ MAA Online How Euler did it, Ed.Sandifer
[2]-Basel Problem @ Wikipedia Basel Problem
[3]-Telescoping Series @ Wikipedia Telescoping Series
[4]-Polygonal Numbers @ Wikipedia Polygonal Number