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2. geometric series - Prove that 1/2 + 1/4 + 1/8 ....... = 1 -...

   math.stackexchange.com/questions/2777959/prove-that-1-2-1-4-1-8-1

   I've often heard that instead of adding up to a little less than one, 1/2 + 1/4 + 1/8... = 1. Is there any way to prove this using equations without using Sigma, or is it just an accepted fact?

3. What is the sum of $1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32\\dots$?

   math.stackexchange.com/questions/2747719/what-is-the-sum-of-1-1-2-1-4-1-8-1-16...

   I couldn't find out the exact value of this operation. $$1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32 \dots$$ You go 1 units right on the number line, half of it to the left, half of the previous one to the

4. Sequence Sum {1/2 + 1/4 + 1/6 +...} to infinite

   math.stackexchange.com/questions/193515/sequence-sum-1-2-1-4-1-6-to-infinite

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5. Sum of 1 + 1/2 - 1/3 +.... + 1/n - Mathematics Stack Exchange

   math.stackexchange.com/questions/3367037/sum-of-1-1-2-1-3-1-n

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6. what is 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 +1/9

   math.stackexchange.com/questions/3454007

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7. Does $ 1 + 1/2 - 1/3 + 1/4 +1/5 - 1/6 + 1/7 + 1/8 - 1/9 + ...$...

   math.stackexchange.com/questions/1699752

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8. Sum of Infinite Series $1 + 1/2 + 1/4 + 1/16 + \\cdots$

   math.stackexchange.com/questions/583472/sum-of-infinite-series-1-1-2-1-4-1-16...

   Everyone knows about the classic $$ \sum_{i=1}^{\infty} \dfrac{1}{2^i} = 1 $$ However, is there any way to find $$ \sum_{i=0}^{\infty} \dfrac{1}{2^{2^i}} = \dfrac12 ...

9. The sum of series $1-1/2-1/4+1/3-1/6-1/8+1/5-....$?

   math.stackexchange.com/questions/2096604/the-sum-of-series-1-1-2-1-41-3-1-6-1-81-5

   I know that the sum of series $1-1/2+1/3-1/4+1/5..= log 2.$ And we can see that by rearranging the terms of the series given in question, i would land on the series as above. so this should mean th...

10. Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \\cdots$

    math.stackexchange.com/questions/2967089/compute-1-1-3-1-2-1-5-1-7-1-4-1-9-cdots

    It seems to me the pattern is you are adding the following triplets: $\frac 1 {4k+1} +\frac 1 {4k+3}-\frac 1 {2^{k+1}} $. If this converged we could rearrange the terms. The infinite sum of $\sum\frac {-1} {2^{k+1}}$ is $-1$ which is finite so that would mean the sum $\sum (\frac 1 {4k+1}+\frac 1 {4k+3})= \sum \frac 1 {2k+1} $ is finite.

11. What's the limit of $\\prod_1^\\infty...

    math.stackexchange.com/questions/491948

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