{"id":281,"date":"2013-05-30T22:12:24","date_gmt":"2013-05-31T02:12:24","guid":{"rendered":"http:\/\/www.curtisbright.com\/bln\/?p=281"},"modified":"2013-05-30T22:12:24","modified_gmt":"2013-05-31T02:12:24","slug":"the-intended-cute-solution","status":"publish","type":"post","link":"http:\/\/localhost\/blog\/index.php\/2013\/05\/30\/the-intended-cute-solution\/","title":{"rendered":"The intended cute solution"},"content":{"rendered":"

The question I meant to ask on Sunday was whether $\\sum_{m,n=1}^\\infty1\/(m^2+n^2)^2$ converges or diverges, so I’ll give my solution to that problem now. In fact, the methodology closely resembles the divergence proof<\/a> I gave for the alternate sum, even though this sum converges.<\/p>\n

Since all terms in the sum are positive, the sum either converges absolutely or diverges (the sum\u00a0cannot be\u00a0conditionally convergent<\/a>). In either case the terms of the sum may be rearranged without affecting the convergence. For suppose the sum diverges but converges for some rearrangement: since the terms of the rearranged sum are still positive, the rearranged sum would have to converge absolutely, and therefore converge for all rearrangements, contradicting the supposed diverging arrangement.<\/p>\n

Therefore we can rearrange the terms as we please; in particular, we can sort the terms in decreasing order, which has the effect of grouping together all terms of the form $1\/k^2$ where $k$ is a sum of two squares. The term $1\/k^2$ will appear in the sum once for each solution of $m^2+n^2=k$ in positive integers $m$, $n$.1<\/a><\/sup><\/p>\n

For our purposes it is sufficient to note there are at most $\\sqrt{k}$ solutions to $m^2+n^2=k$. This follows since we know that $1\\leq m\\leq\\sqrt{k}$ and for each value of $m$ there is at most one solution; the only possible value of $n$ which could work is $\\sqrt{k-m^2}$.<\/p>\n

The argument proceeds as follows:<\/p>\n

\\[ \\newcommand{\\N}{\\mathbb{N}}\\begin{align*}
\n\\sum_{m,n\\in\\N}\\frac{1}{(m^2+n^2)^2} &= \\sum_{k=2}^\\infty\\sum_{\\substack{m,n\\in\\N\\\\m^2+n^2=k}}\\frac{1}{k^2} \\\\
\n&\\leq \\sum_{k=2}^\\infty\\frac{\\sqrt{k}}{k^2} \\\\
\n&= \\sum_{k=2}^\\infty\\frac{1}{k^{3\/2}} \\\\
\n&= \\zeta(3\/2)-1
\n\\end{align*} \\]<\/p>\n

The final sum converges since it is a $p$-series<\/a>\u00a0(truncated). By the comparison test, the sum in question also converges.<\/p>\n

\n
<\/div>\n
    \n
  1. The exact number of solutions to $m^2+n^2=k$ is essentially the\u00a0sum of squares function<\/a>\u00a0$r_2(k)$, although since we are exclusively working with solutions in\u00a0positive<\/em> numbers the actual number of solutions will be $r_2(k)\/4$ if $k$ is not a perfect square and $(r_2(k)-4)\/4$ if $k$ is a perfect square.\n

    Via MathWorld, we see that if $r_2(k)$ is nonzero then it is equal to $4B(k)$, where $B(k)$ is the number of divisors of $k$ solely comprised of primes congruent to $1$ mod $4$. In any case, this shows that the maximum number of times $1\/k^2$ can appear in the summation is $d(k)$, the\u00a0number of divisors<\/a> of $k$. In Apostol<\/a>\u00a0(page 296) it is shown that $d(k)=o(k^\\epsilon)$ for any $\\epsilon>0$, so this is a fairly slowly growing function. ↩<\/a><\/span><\/li>\n<\/ol>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

    The question I meant to ask on Sunday was whether $\\sum_{m,n=1}^\\infty1\/(m^2+n^2)^2$ converges or diverges, so I’ll give my solution to that problem now. In fact, the methodology closely resembles the divergence proof I gave for the alternate sum, even though … 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\/281"}],"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=281"}],"version-history":[{"count":0,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts\/281\/revisions"}],"wp:attachment":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/media?parent=281"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/categories?post=281"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/tags?post=281"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}