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Lets assume we have the following summation... and we are asked to evaluate the summation

$$\sum_{i=0}^{\log (n)} (4^i)$$

I know that this is a geometric series and it converges if 4 is less than one. Obviously this is false, therefore the summation cannot converge to a single value. Therefore the result is infinity

Is this the proper approach to solving the summation. Did i even get the correct answer?

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    $\begingroup$ "if 4 is less than one" - that's quite an interesting way to put it. On a more serious note, what is logn? $\endgroup$ – Yuriy S Jan 26 '18 at 8:11
  • $\begingroup$ Sorry i meant to put log(n) $\endgroup$ – Soon_to_be_code_master Jan 26 '18 at 8:12
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    $\begingroup$ @Soon_to_be_code_master You can just use the formula to find the finite sum $\frac{a(r^n-1)}{r-1}$, and then substitute $\log n$ into that. $\endgroup$ – Toby Mak Jan 26 '18 at 8:15
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Notice that this is a finite sum.

The following formula might help you, if $m \in \mathbb{N}$,

$$\sum_{k=0}^{m-1}ar^k=a\left(\frac{1-r^m}{1-r} \right)$$

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  • $\begingroup$ In the OP's case $m \notin \mathbb{N}$, at least in the general case. $\endgroup$ – Yuriy S Jan 26 '18 at 8:18
  • $\begingroup$ Yes, so he will have to modify things. first, we have to figure out what does it mean when the upper limit is not a natural number. $\endgroup$ – Siong Thye Goh Jan 26 '18 at 8:21
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For any $n$, this is a sum of a finite number of terms, so the sum is finite.

Just use the standard formula for the sum of a geometric series and put in 4 as the ratio and log n and 0 as the upper and lower limits.

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  • $\begingroup$ Is there a proof or does it work by definition? (Going from a positive integer limit to a real one) $\endgroup$ – Yuriy S Jan 26 '18 at 8:21
  • $\begingroup$ In a summation where a limit is not an integer, generally the integer part is used. $\endgroup$ – marty cohen Jan 26 '18 at 14:47

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