{"id":264,"date":"2013-05-28T23:57:14","date_gmt":"2013-05-29T03:57:14","guid":{"rendered":"http:\/\/www.curtisbright.com\/bln\/?p=264"},"modified":"2013-05-28T23:57:14","modified_gmt":"2013-05-29T03:57:14","slug":"another-cute-solution","status":"publish","type":"post","link":"http:\/\/localhost\/blog\/index.php\/2013\/05\/28\/another-cute-solution\/","title":{"rendered":"Another cute solution"},"content":{"rendered":"

Previously<\/a> I asked if the summation $\\sum_{m,n=1}^\\infty1\/(m^2+n^2)$ converges or diverges. Actually, the I intended the denominator of the summation term to be $(m^2+n^2)^2$. But never mind, let’s solve the problem as given. To do this, we’ll employ the comparison test<\/a>.<\/p>\n

First, note that<\/p>\n

\\[ m^2+n^2 \\leq m^2+2mn+n^2 = (m+n)^2 , \\]<\/p>\n

so $1\/(m^2+n^2)\\geq1\/(m+n)^2$.<\/p>\n

Next, assume the sum converges; since all terms are positive the sum absolutely converges<\/a> and the terms may be rearranged without affecting its value. In particular, we rearrange the terms in decreasing order, by grouping all terms equal to $1\/k^2$ together for each possible value of $k$.<\/p>\n

Finally, we use the fact that there are exactly $k-1$ solutions to $m+n=k$ in positive integers $m$, $n$. Putting it all together, the argument goes as follows:<\/p>\n

\\begin{align*}
\n\\newcommand{\\N}{\\mathbb{N}}
\n\\sum_{m,n\\in\\N}\\frac{1}{m^2+n^2} &\\geq \\sum_{m,n\\in\\N}\\frac{1}{(m+n)^2} \\\\
\n&= \\sum_{k=2}^\\infty\\sum_{\\substack{m,n\\in\\N\\\\m+n=k}}\\frac{1}{k^2} \\\\
\n&= \\sum_{k=2}^\\infty\\frac{k-1}{k^2} \\\\
\n&\\geq \\frac{1}{2}\\sum_{k=2}^\\infty\\frac{1}{k} \\\\
\n&= \\infty
\n\\end{align*}<\/p>\n

The final inequality just uses $k-1\\geq k\/2$ for $k\\geq2$ and we are left with a harmonic series<\/a>, which diverges. By the comparison test the original sum diverges; this contradicts the assumption that it converges, so the sum really does diverge, as required.<\/p>\n

Next up, I’ll show the question I intended<\/em> to ask: does\u00a0$\\sum_{m,n=1}^\\infty1\/(m^2+n^2)^2$ converge or diverge?<\/p>\n","protected":false},"excerpt":{"rendered":"

Previously I asked if the summation $\\sum_{m,n=1}^\\infty1\/(m^2+n^2)$ converges or diverges. Actually, the I intended the denominator of the summation term to be $(m^2+n^2)^2$. But never mind, let’s solve the problem as given. To do this, we’ll employ the comparison test. … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"_links":{"self":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts\/264"}],"collection":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/comments?post=264"}],"version-history":[{"count":0,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts\/264\/revisions"}],"wp:attachment":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/media?parent=264"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/categories?post=264"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/tags?post=264"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}