meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | 23 | |
| visits | member for | 1 year, 10 months |
| seen | 8 hours ago | |
| stats | profile views | 662 |
Freshman Graduate Student in Mathematics
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Feb 2 |
reviewed | Excellent Logic about systems? |
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Feb 1 |
revised |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Added the link to your result so that people can understand the result directly instead of hunting for it in the comments. :-) |
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Feb 1 |
suggested | suggested edit on Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ |
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Feb 1 |
comment |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Yes. I am Jayesh. Name changed for a month. |
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Feb 1 |
accepted | Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ |
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Feb 1 |
comment |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Nice solution Chris'ssister! :-) Your question and the corresponding answers provide a decent tool. :-) |
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Feb 1 |
awarded | Self-Learner |
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Feb 1 |
revised |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ added 44 characters in body |
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Feb 1 |
revised |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ deleted 201 characters in body |
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Feb 1 |
comment |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Thanks. A different solution. :-) |
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Feb 1 |
revised |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ added 155 characters in body |
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Feb 1 |
comment |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Okay. No problems. :-) Generally, when people downvote, and the OP asks the reason, it is expected that the person who downvoted leave a comment about it. And hence, my assumption. |
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Feb 1 |
revised |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ deleted 204 characters in body |
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Feb 1 |
comment |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Arrrrgh. Thanks. |
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Feb 1 |
comment |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ @ToddWilcox Counterpoint. meta.math.stackexchange.com/q/2244/14082 |
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Feb 1 |
comment |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Added the explanation. I hope its correct. |
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Feb 1 |
revised |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ added explanation for solution of differential equation. |
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Feb 1 |
comment |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ A few reasons 1. It gives me oppurtunity to verify that the solution is indeed correct. (I self study, so even though I get an answer and I am pretty sure about it, there is no real way to verify the solution completely.) 2. It allows probably other people to offer me better solutions. 3. I can do so on a blog, but then it might not get the same attention on the blog. 4. MSE's editing capabilities are better than almost all other blogging software I have found. |
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Feb 1 |
comment |
Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Is the downvote because of self-answering? |
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Feb 1 |
asked | Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ |
