# Convergence of $\sum_{n=1}^{\infty}(-1)^n\frac{n}{f_n}$ where $f_n$ is the $n$'th Fibonacci number

Can we show convergence of$$B=\sum_{n=1}^{\infty}(-1)^n\frac{n}{f_n}$$where $f_n$ is the $n$'th Fibonacci number?

And then can we determine the exact value of $B$?

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It may be useful to know, for this or similar problems, that $f_n$ grows exponentially for large $n$, particularly $f_n\sim\phi^n$ where $\phi=(1+\sqrt{5})/2$. – Jonathan Jan 19 '13 at 4:09

You can conclude it based on alternating series test i.e. if $a(n)$ is an eventually monotone decreasing sequence, converging to $0$, then $$\sum_{n=1}^{\infty} (-1)^n a(n)$$ converges.

In your case, $a(n) = \dfrac{n}{f_n}$ is a monotone decreasing sequence for some $n>m$, converging to $0$, since $f_n$ grows exponentially.

Hence, we have that $$\sum_{n=1}^N (-1)^n\dfrac{n}{f_n}$$ converges.

EDIT Mathematica gives the sum of the infinite series as $\approx 0.2692050394$ and the inverse symbolic calculator seems to suggest this number is not related to any of the other "well-known constants".

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I have a question. Could I apply the same solution replacing $f_n$ with $p_n$, where $p_n$ is the $n$'th prime? – sunny Jan 19 '13 at 3:58
Marvis, $p_n$ is roughly $n\log n$. – Gerry Myerson Jan 19 '13 at 4:02
Also, $n/f_n$ is not monotone decreasing, not if you take $f_1=f_2=1$ as is commonly done. – Gerry Myerson Jan 19 '13 at 4:03
@GerryMyerson Oops what a blunder I made! Thanks for pointing it out! Have removed it. – user17762 Jan 19 '13 at 4:04
Finally, isn't the ordinary everyday alternating series test good enough for the original question (after you clear out the non-decreasing part)? – Gerry Myerson Jan 19 '13 at 4:05