document.write( "Question 1183560: Does the series\r
\n" ); document.write( "\n" ); document.write( "(1/√(n^2+1^2) + 2/√(n^2+2^2) +...+(n-1)/√(n^2+(n-1)^2)+ n/√(n^2+n^2))/n\r
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Algebra.Com's Answer #813954 by ikleyn(52794)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "            This problem is not for beginning Calculus students. \r
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\n" ); document.write( "\n" ); document.write( "            Therefore, I will give my solution without going into details,  assuming that the reader has an adequate level.\r
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document.write( "The sum is\r\n" );
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document.write( "    \"S%5Bn%5D\" = \"%281%2Fn%29%2Asum%28k%2Fsqrt%28n%5E2%2Bk%5E2%29%2C+k=1%2Cn%29\".      (1)\r\n" );
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document.write( "Each term under the sum symbol can be estimated this way\r\n" );
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document.write( "    \"k%2Fsqrt%28n%5E2%2Bk%5E2%29\" <= \"k%2Fsqrt%28n%5E2%29\" = \"k%2Fn\"   from the top,                   (2)\r\n" );
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document.write( "and\r\n" );
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document.write( "    \"k%2Fsqrt%28n%5E2%2Bk%5E2%29\" >= \"k%2Fsqrt%28n%5E2%2Bn%5E2%29\" = \"k%2F%28n%2Asqrt%282%29%29\"  from the bottom,       (3)\r\n" );
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document.write( "or  \r\n" );
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document.write( "    \"%281%2Fsqrt%282%29%29%2A%28k%2Fn%29\" <= \"k%2Fsqrt%28n%5E2%2Bn%5E2%29\" <= \"k%2Fn\".      (4)\r\n" );
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document.write( "It gives, in turn, the following estimations for the entire sums  \"S%5Bn%5D\"\r\n" );
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document.write( "    \"%281%2Fsqrt%282%29%29%2A%281%2Fn%29%2Asum%28k%2Fn%2Ck=1%2Cn%29\" <= \"S%5Bn%5D\" <= \"%281%2Fn%29%2Asum%28k%2Fn%2Ck=1%2Cn%29\".      (5)\r\n" );
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document.write( "Next, each sum  \"sum%28%28k%2Fn%29%2Ck=1%2Cn%29\"  contains the sum of arithmetic progression  1 + 2 + 3 + . . . + n = \"%28n%2A%28n%2B1%29%29%2F2\",\r\n" );
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document.write( "therefore,   \"sum%28%28k%2Fn%29%2Ck=1%2Cn%29\" = \"%28n%2A%28n%2B1%29%29%2F%282%2An%29\" = \"%28n%2B1%29%2F2\".\r\n" );
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document.write( "From this point, estimations (5) can be re-written\r\n" );
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document.write( "    \"%281%2Fsqrt%282%29%29%2A%281%2F2%29\" <= \"S%5Bn%5D\" <= \"%28%28n%2B1%29%2F%282n%29%29\".      (6)\r\n" );
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document.write( "Thus the terms  \"S%5Bn%5D\"  are asymptotically between  \"1%2F%282%2Asqrt%282%29%29\"  and  \"1%2F2\".\r\n" );
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document.write( "        The sequence  \"S%5Bn%5D\"  is, actually,  monotonically DECREASING  sequence.\r\n" );
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document.write( "From the first glance, it seems to be unexpected statement, but it is true and it can be strictly established formally.\r\n" );
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document.write( "Therefore, of the two estimations (6) from the top and from the bottom, the only BOTTOM estimation makes sense for us now.\r\n" );
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document.write( "So, we have the monotonically decreasing sequence  \"S%5Bn%5D\"  limited by the value  \"1%2F%282%2Asqrt%282%29%29\"  from the bottom --- THEREFORE,\r\n" );
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document.write( "     +--------------------------------------------+\r\n" );
document.write( "     |   the sequence  \"S%5Bn%5D\"  is converged.         |\r\n" );
document.write( "     +--------------------------------------------+\r\n" );
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