{"id":1183,"date":"2015-03-31T20:27:07","date_gmt":"2015-04-01T00:27:07","guid":{"rendered":"http:\/\/www.curtisbright.com\/bln\/?p=1183"},"modified":"2015-03-31T20:27:07","modified_gmt":"2015-04-01T00:27:07","slug":"the-value-of-the-trigonometric-harmonic-series-revisited","status":"publish","type":"post","link":"http:\/\/localhost\/blog\/index.php\/2015\/03\/31\/the-value-of-the-trigonometric-harmonic-series-revisited\/","title":{"rendered":"The value of the trigonometric harmonic series revisited"},"content":{"rendered":"<p>Shortly after my <a href=\"http:\/\/www.curtisbright.com\/bln\/2015\/02\/28\/the-value-of-the-trigonometric-harmonic-series\/\">last post<\/a>\u00a0I realized there was a simpler way of determining the exact value of the series $\\sum_{n=1}^\\infty\\cos n\/n$. Instead of following\u00a0the method I previously described which required the intricate analysis of some integrals, one can simply use\u00a0the formula<\/p>\n<p>\\[ \\sum_{n=1}^\\infty\\frac{a^n}{n} = -\\ln(1-a) \\]<\/p>\n<p>which is valid for $a\\in\\mathbb{C}$ which satisfies $\\lvert a\\rvert\\leq 1$ and $a\\neq 1$. This comes from a simple rewriting\u00a0of the so-called\u00a0<a href=\"http:\/\/en.wikipedia.org\/wiki\/Mercator_series#Complex_series\">Mercator series<\/a>\u00a0(replace $x$ with $-x$ in the Taylor series of $\\ln(1+x)$ and then take the negative).<\/p>\n<p>Then we have<\/p>\n<p>\\begin{align*}<br \/>\n\\sum_{n=1}^\\infty\\frac{\\cos n}{n} &amp;= \\sum_{n=1}^\\infty\\frac{e^{in}+e^{-in}}{2n} \\\\<br \/>\n&amp;= -\\bigl(\\ln(1-e^i)+\\ln(1-e^{-i})\\bigr)\/2 \\\\<br \/>\n&amp;= -\\ln\\bigl((1-e^i)(1-e^{-i})\\bigr)\/2 \\\\<br \/>\n&amp;= -\\ln(2-e^i-e^{-i})\/2 \\\\<br \/>\n&amp;= -\\ln(2-2\\cos1)\/2 \\\\<br \/>\n&amp;\\approx\u00a00.0420195<br \/>\n\\end{align*}<\/p>\n<p>since $\\lvert e^i\\rvert=\\lvert e^{-i}\\rvert=1$, but $e^i\\neq1$ and $e^{-i}\\neq1$.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Shortly after my last post\u00a0I realized there was a simpler way of determining the exact value of the series $\\sum_{n=1}^\\infty\\cos n\/n$. Instead of following\u00a0the method I previously described which required the intricate analysis of some integrals, one can simply use\u00a0the &hellip; <a href=\"http:\/\/localhost\/blog\/index.php\/2015\/03\/31\/the-value-of-the-trigonometric-harmonic-series-revisited\/\">Continue reading <span class=\"meta-nav\">&rarr;<\/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\/1183"}],"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=1183"}],"version-history":[{"count":0,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts\/1183\/revisions"}],"wp:attachment":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/media?parent=1183"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/categories?post=1183"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/tags?post=1183"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}