converting slow convergence to fast one. I have to find sum of the following infinite series
    ∞  

ɸ(x) =∑ 1/ k(k+x)        for x = 0,0.1 ,0.2,……….1.0
    k=1 

I solved it like this
for x= 1   ∞ 
           ∑ 1/ k(k+x)        =∑ 1 /k^2 - (1/(k^2)(k+1))
           k=1
                              = ∑ 1 /k^2 - ((1/(k^2)) - (1/ k (k+1)))
                              = ∑ 1 /k^2 - ((1/(k^2)) - ((1/ k ) - (1/(k+1))))
                              = ∑ 1/ k^2 - ( 1/k ^2) - 1/k + 1/ (k+1)
                              = ∑ 1/ k ^2  -1 / k ^2 +1/k- 1 /(k+1)
                              = ∑ 1/ k  - 1 / (k+1)
This is known to be equal to 1
When I write Matlab code to find its sum with error less than 0.5 *10 power -8
 I get required result BUT the series converges very slowly. i.e. it converges for taking 80000000 terms with time approximately 15 seconds.
Does anyone have idea, how to accelerate the series ? 
Any help will be appreciated. Thanks.
 A: The closed-form solution is $\phi(x) = \dfrac{\gamma+\Psi(x+1)}{x}$ where $\Psi$ is the digamma function (psi) in Matlab) and $\gamma$ is Euler's constant, approximately
$0.5772156649015329$.
As for speeding it up, you could try an Euler-Maclaurin series.
EDIT: By "fourth degree", I suppose you mean you want to use  a sum of $N$ terms and get an error $O(1/N^4)$.  Indeed, it is not hard to do  that. Rather than using Euler-Maclaurin, I'll use explicit estimates. One way is related to the   Midpoint Rule approximation of integrals by sums, but turned into an
approximation of sums by integrals.
If $f$ is a smooth function, and $n$ is given, let 
$$L(h) = \int_{n - h}^{n+h} f(t)\; dt - 2 h f(n)$$
Note that $L(0) = L'(0) = 0$ and $$L''(h) = f'(n+h) - f'(n-h) = \int_{n-h}^{n+h} f''(t)\; dt$$
If we have a bound $u \le f''(t) \le v$ for $n-h \le t \le n+h$, then
$2u h \le L''(h) \le 2 v h$.  Integrating twice, $u h^3/3 \le L(h) \le v h^3/3$.
In your case you want $f(t) = 1/(t(t+x))$ so $$f''(t) = \dfrac{2}{t^3(t+x)} + \dfrac{2}{t^2(t+x)^2} + \dfrac{2}{t(t+x)^3}$$  Note that for $t > 0$ and $x \ge 0$ this is a decreasing function of $t$ (because each term is decreasing).
Thus for $n > h \ge 0$, we can take $u = f''(n+h)$ and $v = f''(n-h)$.  In particular, with $h = 1/2$, we get
$$ \dfrac{f''(n+1/2)}{24} \le L(1/2) = \int_{n-1/2}^{n+1/2} f(t)\; dt - f(n) \le \dfrac{f''(n-1/2)}{24} $$
Summing for $n$ from $N$ to $\infty$ and turning it around, 
$$ \int_{N-1/2}^\infty f(t)\; dt - \sum_{n=N}^\infty \dfrac{f''(n+1/2)}{24}
\ge \sum_{n=N}^\infty f(n) \ge \int_{N-1/2}^\infty f(t)\; dt - \sum_{n=N}^\infty \dfrac{f''(n-1/2)}{24}$$
Again, since $f''$ is decreasing, we can estimate the sums of $f''$ by integrals:
$$\eqalign{ \int_{N-1/2}^\infty f(t)\; dt + f'(N+1/2) &= \int_{N-1/2}^\infty f(t)\; dt - \int_{N+1/2}^\infty \dfrac{f''(t)}{24}\; dt\cr
&\ge \sum_{n=N}^\infty f(n) \cr &\ge \int_{N-1/2}^\infty f(t)\; dt - \int_{N-3/2}^\infty \dfrac{f''(t)}{24}\; dt = \int_{N-1/2}^\infty f(t)\; dt + f'(N-3/2)\cr}$$
The difference between these upper and lower bounds is
$$ f'(N+1/2) - f'(N-3/2) $$
In your case note that this is $O(1/N^4)$.
A: Note that the $n$-th term
$$
T_n =\frac{1}{n(n+x)}$$
and
$$
\frac{T_{n+1}}{T_n} \to 1 ~~\text{as}~~ n \to \infty
$$
So the ratio test says that the convergence is slower than a geometric sequence. So expect very slow convergence and not much you can do about it!
By the way the case $x=0$ is called the Basel problem and was solved by Euler who showed that
$$
\sum \frac{1}{k^2} = \frac{\pi^2}{6}
$$
I don't think you can speed it up using elementary means.
A: Not surprizing, Robert Israel gave you the exact formula for your problem. On the other hand, user44197 showed some interesting features of the function.  
When you plot the function for $0<x<1$, it is not far away from a straight line (small curvature) and, innocently (at my ages !), I thought that some Taylor expansions (as suggested by Robert Israel) could be quite efficient. Surprizingly , they are not.   
It is sure that, with some work, it could be possible to find closed form approximation of the function for this restricted range of $x$. But, I am afraid we should never be able to reach the level of accuracy you require.
