# Thread: Sum to infinity of log n/n! series

1. ## Sum to infinity of log n/n! series

Hello,

I require the value of the following sum for some complexity analysis. I would be grateful if someone can help me out. Even a valid upper bound is okay.

$\sum_{i=2}^{\infty} \frac{\log i}{(i-2)!}$

Pardon me if I have done/said anything wrong.

Abhiram.

2. I'm assuming you're using log base e: ln. Well, $\ln(i)\le i\;\forall\,i> 0$. Hence,

$
\sum_{i=2}^{\infty} \frac{\log(i)}{(i-2)!}\le\sum_{i=2}^{\infty} \frac{i}{(i-2)!}=3e.
$

3. Oh well, yes. That is a valid bound. I was hoping for a tighter one. And it is $\log_2$. Empirically I find that it converges to 4.10.

4. Ok, instead of using $\log_{2}(i)\le i$, we can use a different slope line through the origin. You can show that the line

$y=\frac{1}{2^{1/\ln(2)}\ln(2)}\,x\ge\log_{2}(x)\,\forall\,x>0$. This is the line through the origin that just touches the $\log_{2}(x)$ graph. The bound is now

$3e\,\frac{1}{2^{1/\ln(2)}\ln(2)}\le 4.3281$.

You could probably do even better if you were consider lines through the point (0,5) that are tangent to the graph of $\log_{2}(x)$.

5. That helps. Thanks. I just wanted to know whether there exists an absolute sum. I guess not.

6. Originally Posted by abhiramn
Hello,

I require the value of the following sum for some complexity analysis. I would be grateful if someone can help me out. Even a valid upper bound is okay.

$\sum_{i=2}^{\infty} \frac{\log i}{(i-2)!}$

Pardon me if I have done/said anything wrong.

Abhiram.
This thing converges so fast that you do not need many terms to get a good estimate:

$\sum_{i=2}^{10} \frac{\log( i)}{(i-2)!}=2.847400..$

and it is fairly easy to bound the error on this

$
\text{error}=\sum_{i=11}^{\infty} \frac{\log (i)}{(i-2)!} < \frac{\log(11)}{9}\sum_{i=11}^{\infty} \frac{1}{(i-3)!}$
$=\frac{\log(11)}{9}\left(e-\sum_{i=3}^{10} \frac{1}{(i-3)!}\right) \approx 7.4229 \times 10^{-6}
$

You can also obtain a lower bound on the error by observing that

$\text{error}=\sum_{i=11}^{\infty} \frac{\log (i)}{(i-2)!} > \log(11)\sum_{i=11}^{\infty} \frac{1}{(i-2)!}\approx 7.3342 \times 10^{-6}$

CB

7. CB: are you using $\log_{2}(i)$? Or is this a lower bound? The number you came up with seems a bit low to me, at least given the numerical approximation to the sum that Mathematica gives me.

8. Originally Posted by Ackbeet
CB: are you using $\log_{2}(i)$? Or is this a lower bound? The number you came up with seems a bit low to me, at least given the numerical approximation to the sum that Mathematica gives me.
No $\log_e$ but if you intended $\log_2$ in the original there is a simple scale factor to get that from what is in my post, since $\log_e(x)=\log_2(x)/\log_2(e)$. So if base 2 was intended multiply through by $\log_2(e)$

CB

9. Got it. Yeah, in post #3, the OP-er mentions it's log base 2.