{"id":498,"date":"2014-04-29T04:36:08","date_gmt":"2014-04-29T08:36:08","guid":{"rendered":"http:\/\/www.curtisbright.com\/bln\/?p=498"},"modified":"2014-04-29T04:36:08","modified_gmt":"2014-04-29T08:36:08","slug":"harmonic-series-variant-solution","status":"publish","type":"post","link":"http:\/\/localhost\/blog\/index.php\/2014\/04\/29\/harmonic-series-variant-solution\/","title":{"rendered":"Harmonic series variant solution"},"content":{"rendered":"

Last year I asked<\/a> if the series $\\sum_{n=1}^\\infty\\cos n\/n$ converges or diverges. Although the series looks rather simple, the standard convergence tests learned in Calculus 1 do not apply directly. However, there is a trick which allows one to rewrite the series into a form where the standard tests can be used.<\/p>\n

The trick is to use summation by parts<\/a>, the discrete analog of integration by parts<\/a>. Although less well-known, summation by parts does have its uses; for example, it was the method of proving Abel’s summation formula which has come up\u00a0before<\/a>.<\/p>\n

To simplify the exposition, it is convenient to define a function for the partial sums of the series’ numerators:<\/p>\n

\\[ C(m) := \\sum_{n=1}^m \\cos n \\]<\/p>\n

Now, using summation by parts the series may be rewritten so that<\/p>\n

\\[ \\sum_{n=1}^m\\frac{\\cos n}{n} = \\frac{C(m)}{m} + \\sum_{n=1}^{m-1} C(n)\\Bigl(\\frac{1}{n}-\\frac{1}{n+1}\\Bigr) . \\tag{1} \\]<\/p>\n

The key observation to make now is that $C(m)$ is bounded. This can be seen by using the complex exponential expression of the cosine, and summing a geometric series to find a closed-form expression for $C(m)$, as follows:<\/p>\n

\\begin{align*}
\nC(m) &= \\sum_{n=1}^m\\frac{e^{in}+e^{-in}}{2} \\\\
\n&= \\frac{e^i}{2}\\Bigl(\\frac{e^{im}-1}{e^i-1}\\Bigr)+\\frac{e^{-i}}{2}\\Bigl(\\frac{e^{-im}-1}{e^{-i}-1}\\Bigr) \\\\
\n&= \\frac{(e^{im}-1)(1-e^i)+(e^{-im}-1)(1-e^{-i})}{2(e^i-1)(e^{-i}-1)} \\\\
\n&= \\frac{(e^{im}+e^{-im})-(e^{i(m+1)}+e^{-i(m+1)})}{2(2-(e^i+e^{-i}))}-\\frac{1}{2} \\\\
\n&= \\frac{\\cos m-\\cos(m+1)}{2(1-\\cos1)}-\\frac{1}{2}
\n\\end{align*}<\/p>\n

By the triangle inequality, this has an absolute upper bound of<\/p>\n

\\[ \\lvert C(m)\\rvert \\leq \\frac{1}{1-\\cos1}+\\frac{1}{2}<3 . \\]<\/p>\n

From this it follows that $C(m)\/m\\to0$ as $m\\to\\infty$, and after taking the limit (1) becomes<\/p>\n

\\[ \\sum_{n=1}^\\infty\\frac{\\cos n}{n} = \\sum_{n=1}^\\infty{}C(n)\\Bigl(\\frac{1}{n}-\\frac{1}{n+1}\\Bigr) ,\u00a0\\tag{2}\u00a0\\]<\/p>\n

and we can use the absolute comparison test on the right sum. Using the fact that<\/p>\n

\\[ \\frac{1}{n}-\\frac{1}{n+1} = \\frac{1}{n^2+n} \\leq \\frac{1}{n^2}\u00a0\\]<\/p>\n

we have $\\bigl\\lvert C(n)(\\frac{1}{n}-\\frac{1}{n+1})\\bigr\\rvert<3\/n^2$, and $\\sum_{n=1}^\\infty 3\/n^2=3\\zeta(2)$ is absolutely convergent, so by the comparison test the right sum in (2) is absolutely convergent as well. Thus the left side of (2) must converge, so $\\sum_{n=1}^\\infty\\cos n\/n$ converges.<\/p>\n","protected":false},"excerpt":{"rendered":"

Last year I asked if the series $\\sum_{n=1}^\\infty\\cos n\/n$ converges or diverges. Although the series looks rather simple, the standard convergence tests learned in Calculus 1 do not apply directly. However, there is a trick which allows one to rewrite … 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\/498"}],"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=498"}],"version-history":[{"count":0,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts\/498\/revisions"}],"wp:attachment":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/media?parent=498"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/categories?post=498"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/tags?post=498"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}