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Here is an interesting infinite series. It would be great to see a method to evaluate it, if possible. I know it converges to a little less than 11/40

$\displaystyle\sum_{k=1}^{\infty}\frac{1}{4^{k}+k!}$

I could not think of any good identities to start this.

Thanks a million to those who can show how to evaluate it.

Maybe even in general, $\displaystyle\sum_{k=1}^{\infty}\frac{1}{x^{k}+k!}$, where $x\geq 1$

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This looks hopeless to me. $x^k$ and $k!$ are such different beasts. In fact, I would be surprised if it was solvable even after replacing $4^k$ by $1$. –  TonyK Mar 5 '11 at 17:08
    
Inverse symbolic calculator doesn't seem to get any hits. Most usual approaches are not working either. –  Eric Naslund Mar 5 '11 at 17:39
1  
@Tony: Interestingly $$\sum_{k=1}^\infty \frac{1}{1+k!}$$ doesn't get any hits in the inverse calculator, other than itself. (That is, there is an entry specifying exactly that sum). This gives some evidence that there may be no known closed form for either. –  Eric Naslund Mar 5 '11 at 17:58
    
The Only intuition I can give is that this series should converge very fast. So probably evaluating it to the 10th term would give you a nice idea on the limit up to a very small $\varepsilon$. But that much you probably know. –  Asaf Karagila Mar 5 '11 at 20:37
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@approximist The series is less than 4^-k which is known to converge. The issue is only: what does it converge to? –  barrycarter Mar 6 '11 at 19:46
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1 Answer 1

up vote 2 down vote accepted

The only approach I can think of is to narrow down the difference in the remainder term between the exact value of the series and the value of some partial sum that we use as estimate and judge how good or close our estimate is. Take the remainder term $R_n=\sum_{k=n+1}^\infty \frac{1}{4^k+k!}$ and $T_n=\sum_{k=n+1}^\infty \frac{1}{4^k}$ we know that $R_n\lt T_n\lt\int_{n+1}^\infty \frac{1}{4^k}dk $. The higher you choose your $n$ to be the lower the remainder or the error term. For example, $n=8$ partial sum is $s_7=0.2745411421$ and $R_7\lt 0.000011006$. So the partial sum is correct to atleast three or four decimal places. (This was too long for a comment.)

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Thank you very much. I have mistakenly thought in the past that, if a series converges, it must have a closed form. But, like many antiderivatives, that is not the case. Thanks for your input. –  Cody Mar 15 '11 at 9:58
    
@Approximist Your integral over $k$ makes no sense (and an upper bound of $T_n$ would involve an integral starting from $n$, and not from $n+1$). Rather, use $T_n=1/(3\cdot4^n)$. –  Did Mar 15 '11 at 10:21
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