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I got that this one diverged by Integral Test after I ended up using the integral...Mathematica said the same. Anyone agree that it diverges? I just want to make sure I'm not crazy since this problem cost me a lot of time and struggle... --[[User:Reckman|Randy Eckman]] 01:37, 3 November 2008 (UTC) | I got that this one diverged by Integral Test after I ended up using the integral...Mathematica said the same. Anyone agree that it diverges? I just want to make sure I'm not crazy since this problem cost me a lot of time and struggle... --[[User:Reckman|Randy Eckman]] 01:37, 3 November 2008 (UTC) | ||
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| + | Yeah, I got that it diverged as well. So I hope you're not crazy. [[User:Jhunsber|His Awesomeness, Josh Hunsberger]] | ||
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| + | Yep. Not because of the initial increase, though. --[[User:Jmason|John Mason]] | ||
Latest revision as of 11:16, 3 November 2008
Plotting the expression ln(n)/sqrt(n) shows that the function first increases until about x = 7 or 8, and then decreases as x goes to infinity. In order to use the Integral test, however, doesn't the function have to be continually decreasing over the entire domain of the sum? --Randy Eckman 21:38, 2 November 2008 (UTC)
Nah. It just needs to be decreasing as n approaches infinity. Think about it this way: you could change the sum so it includes all of the increasing terms, plus the sum from n past that point to infinity, and it would still be finite. --John Mason
I got that this one diverged by Integral Test after I ended up using the integral...Mathematica said the same. Anyone agree that it diverges? I just want to make sure I'm not crazy since this problem cost me a lot of time and struggle... --Randy Eckman 01:37, 3 November 2008 (UTC)
Yeah, I got that it diverged as well. So I hope you're not crazy. His Awesomeness, Josh Hunsberger
Yep. Not because of the initial increase, though. --John Mason
