| bio | website | 2718.us |
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| location | ||
| age | ||
| visits | member for | 2 years, 10 months |
| seen | 5 hours ago | |
| stats | profile views | 2,626 |
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May 20 |
awarded | Good Answer |
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May 18 |
awarded | Constituent |
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May 16 |
accepted | Solving Triangles (finding missing sides/angles given 3 sides/angles) |
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May 16 |
awarded | Nice Question |
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May 16 |
revised |
Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I hate all-TeX titles. |
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May 16 |
comment |
How do I calculate the new x,y coordinates and width/height of a re-sized group of objects? @asloob: yup, fixed now, thanks. |
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May 16 |
revised |
How do I calculate the new x,y coordinates and width/height of a re-sized group of objects? edited body |
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May 14 |
awarded | Popular Question |
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May 11 |
awarded | Notable Question |
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May 6 |
awarded | Caucus |
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Apr 16 |
comment |
Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ Ahh, right—in my quick read, I hadn't thought about the difference between the limit of the finite series and the infinite series. I hadn't meant to imply that the two sides of that = weren't equal; I more meant that the infinite series related to the left side is conditionally convergent and rearrangement would make it not equal to the right side. |
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Apr 16 |
comment |
Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ Is it the change in form of the series, across the = in the line above (3), that is dependent on the ordering of the series? |
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Apr 7 |
awarded | Good Question |
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Apr 3 |
awarded | Good Question |
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Mar 24 |
awarded | Notable Question |
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Mar 24 |
awarded | Good Answer |
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Mar 22 |
revised |
Proving the Shoelace Method at the Precalculus Level added 8 characters in body |
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Mar 10 |
awarded | Popular Question |
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Feb 24 |
revised |
Why is the volume of a sphere $\frac{4}{3}\pi r^3$? added 66 characters in body |
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Feb 19 |
awarded | Famous Question |