1
$\begingroup$

Let $F_k$ be Fibonacci numbers. I am looking for a closed form of the sum $\sum_{k=1}^\infty F_k/k!$.

I tried to use Wolfram Alpha, but it is not doing the sum Fibonacci[k]/k! , k=1 to infinity.

Can someone tell what is the problem with WA and what this sum equals to?

$\endgroup$
5
$\begingroup$

A hint:

You can prove for yourself or find in a book a formula of the form $${\rm fib}(k)=a\lambda^k + b\mu^k$$ with certain constants $a$, $b$, $\lambda$, $\mu$. The requested sum can then be easily written as a sum of two exponentials.

$\endgroup$
2
$\begingroup$

Im getting the solution: $$ \frac{e^\phi - e^\varphi}{\sqrt{5}} $$

Since it is known that $F_n = \frac{\phi^n - \varphi^n}{\sqrt{5}}$.

$$\sum_{k=0}^\infty \frac{F_k}{k!} = \sum_{k=0}^\infty \frac{\phi^k - \varphi^k}{k!\sqrt{5}} = \frac{1}{\sqrt{5}}\left[\sum_{k=0}^\infty \frac{\phi^k}{k!} - \sum_{k=0}^\infty\frac{\varphi^k}{k!}\right]$$

Youre welcome.

$\endgroup$
  • 1
    $\begingroup$ Im sure the down voting was well-deserved. $\endgroup$ – CogitoErgoCogitoSum Dec 6 '14 at 22:09
  • 1
    $\begingroup$ Why bump an old question? $\endgroup$ – Cyclohexanol. Dec 6 '14 at 22:22
  • 1
    $\begingroup$ I dont check the dates on these questions. It was near the top of the list already when I clicked on it. Besides, how does a legitimate and valid answer deserve a down thumb, regardless of how old it is? $\endgroup$ – CogitoErgoCogitoSum Dec 6 '14 at 22:25
  • 1
    $\begingroup$ I did not downvote, just stating a possible reason. $\endgroup$ – Cyclohexanol. Dec 6 '14 at 22:27
0
$\begingroup$

Using Mma, Limit[Sum[Fibonacci[k]/k!, {k, 1, p}], p -> Infinity] gives an answer and
Sum[Fibonacci[k]/k!, {k, 1, Infinity}] gives the same. Is your problem with Wolfram Alpha ?

$\endgroup$
  • $\begingroup$ computation time out every time. Now please tell the numerical value now. $\endgroup$ – Waqar Ahmad Nov 14 '13 at 12:00
  • $\begingroup$ The numerical value is 2.014322733458315736581346. But I think you should establish the analytical value using what lab bhattacharjee sent you. The formula is quite simple. $\endgroup$ – Claude Leibovici Nov 14 '13 at 12:07
  • $\begingroup$ According to Maple, with: evalf(sum(fibonacci(k)/k!,k=1..infinity)); Giving: 2.014322733. But, does there exist an "exact" solution? Hmm, "the formula is quite simple". I'm quite curious .. $\endgroup$ – Han de Bruijn Nov 14 '13 at 12:09
  • $\begingroup$ @HandeBruijn. Yes and it is (-1 + E**Sqrt(5))/(Sqrt(5)*E**(2/(1 + Sqrt(5)))) $\endgroup$ – Claude Leibovici Nov 14 '13 at 12:10
  • $\begingroup$ @Claude_Leibovici: Numerically the same, OK, but how to prove it? $\endgroup$ – Han de Bruijn Nov 14 '13 at 12:16
-2
$\begingroup$

Solve y'' = y' + y both via power series and then by elementary methods (substitution of exponentials) and setting boundary conditions y(0) = 0, and y'(0) = 1. The result is the following identity:

$\sum_{n=1}^\infty \frac{F_n}{n!} x^n = \frac{2}{\sqrt{5}} e^\frac{x}{2} \sinh(\frac{\sqrt{5}}{2}x)$

If you don't feel like solving the differential equation yourself you can easily verify by substituting each side of the above equation into the differential equation and boundary conditions and then remind yourself via the existence and uniqueness theorem that the two solutions must be the same. Meaning they are the same at the point x=1.

Setting x=1 gives

$\sum_{n=1}^\infty \frac{F_n}{n!} = 2 \sqrt{\frac{e}{5}} \sinh(\frac{\sqrt{5}}{2}) = \frac{e^{\phi_+} - e^{\phi_-}}{\sqrt{5}}$

where $\phi_\pm = \frac{1 \pm \sqrt{5}}{2}$ are the golden ratio and it's congugate. I.e. the two solutions to the quadratic equation $y^2 = y + 1$. (Note that this is the same as the original differential equation with the order of the derivative replaced by the power of y: $y^{(n)} \rightarrow y^n$. Is there a reason for this or is it just an interesting coincidence?)

The numerical value to 1234 decimal places is:

2.0143227334583157365813462554697591356591114695811241821088403766742128397097006637111011319457016312404491456095258793427009006861303485673596801638909160410193717669927023995171048184677900815088560097082872535199871890392172995661754732113194156588018029328948474213043876788216048899918772417664402571617571139330772358975323915258959445759063347851074795879633549994051398333466312388233898626203203269110709326570399986964305691732913351786602533868094239267790638117143928091599301761319009602970110828764189441445889515565022310301247296449293897072832251096862983584326320448058898294430874162784325966040344781269902152316671151601869355616381547811631999794833349897492383827197553543970970099927748899115338963618621987298237221140308693191390458816782367508072038660426129042190625209844154130995652380767395098326152648319830337117521840226032419661899959469674966107384253745331849405956049149602323282511659984227035172708526550586977824786544373977008821755560614931363394267293413160342932047952358802051132151029521708819143029550619292413657149885241166291353201988752249793933504301075491208793989947801376838994380077603990476851019102426649383266999848219352544560978465856984179911553255421975337456890201316269

A fun exercise is to write a C program to generate this (not as easy as you might assume due to the limits of native C data types).

$\endgroup$
  • 1
    $\begingroup$ Simply stating a solution is not helpful in order to understand the problem in more detail. Please add you whole evaluation. $\endgroup$ – mrtaurho Nov 17 '18 at 17:33
  • $\begingroup$ I have added further detail which should satisfy your complaint but I won't give more than that. It is a fun problem which should be solvable by an undergraduate differential equations student and I'm not about to ruin their fun. I'm sure their professor expects them to solve it themselves and not simply copy a solution off the internet. If it were my class I would also require them to write a computer program to calculate the sum of the series to 15 decimal places (i.e. size of double precision floating point). A student getting 200 decimal places from a program written in C would impress me. $\endgroup$ – 1o_o7 Nov 19 '18 at 10:31

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.