Inequality for Fibonacci to find an upper bound of harmonic Fibonacci series I want to find an sharp upper bound for $$\sum_{n=1}^{\infty}\frac{1}{F_n}$$which $F_n $ is the n$th$ term of Fibonacci sequence .
I wrote a Matlab program to find an upper bound ,$\sum_{n=1}^{10^6}\frac{1}{F_n}<4$
  Now my question is:(1):Is there an inequality to find this ?
(2): Is that series have a close form ?
$${F_n} = \frac{{{\varphi ^n} - {{( - \varphi )}^{ - n}}}}{{\sqrt 5 }}\to \\\sum_{n=1}^{\infty}\frac{1}{F_n}=\sum_{n=1}^{\infty}\frac{\sqrt 5}{{{\varphi ^n} - {{( - \varphi )}^{ - n}}}}\\\leq \sum_{n=1}^{\infty}\frac{\sqrt 5}{{{(\frac{{1 + \sqrt 5 }}{2} ) ^n} }}=\frac{\sqrt5}{1-\frac{1}{\frac{{1 + \sqrt 5 }}{2}}}\approx12.18\\$$I am thankful for a hint or solution which can bring a sharper upper bound .
 A: As Jyrki Lahtonen points out in the comments, $\sum_{n=1}^\infty\frac{\sqrt5}{\varphi^n}$ isn't necessarily the right bound, since it doesn't dominate the original series term-by-term.
As for a closed form, this identity is known:
$$\sum_{n=1}^\infty\frac{1}{F_n}=\frac{\sqrt5}{4}\left(\vartheta_2^2(\varphi^{-2}) + \frac{\log5 + 2\psi_{\varphi^{-4}}(1) - 4\psi_{\varphi^{-2}}(1)}{2\log\varphi}\right)$$
where $\vartheta_2(q)$ is the Jacobi theta function at $z=0$, and $\psi_q$ is the $q$-digamma function. See Wikipedia and MathWorld on the "Reciprocal Fibonacci constant", and other Math.SE questions such as What is the sum of Fibonacci reciprocals?
A: Here is a systematic method for computing upper bounds for the sum without much work.
By induction,
if $F_N \ge b^N$ and $b+1 \ge b^2$, then 
$F_n \ge b^n$ for $n \ge N$.
Therefore, 
$$
\sum_{n=1}^\infty {1\over F_n}
\le
\sum_{n=1}^{N-1} {1\over F_n}
+
\sum_{n=N}^\infty {1\over b^n}
=
\sum_{n=1}^{N-1} {1\over F_n}
+
\frac{1}{b^{N-1}(b-1)}
$$
This upper bound gets closer to the actual sum when $b$ gets larger, but then we need larger $N$:
\begin{array}{crl}
b &N &sum \\
1.3 &4 &4.0172204521317 \\
4/3 &5 &3.7825520833333 \\
1.4 &6 &3.4981694135380 \\
1.5 &11 &3.3651521005948 \\
1.6 &72 &3.3598856662432 \\
\end{array}
Since $1.6 \approx \phi$, which is the largest possible $b$, the last value is quite close to the actual value:
$$
3.359885666243177553172011302918927179688905133732\cdots
$$
A: Using Wolfram Mathematica answer is :
Wolfram Mathematica Code

