[et_pb_section fb_built="1" admin_label="Blog Hero" _builder_version="3.22" use_background_color_gradient="on" background_color_gradient_start="rgba(114,114,255,0.24)" background_color_gradient_end="#ffffff" background_blend="multiply" custom_padding="0|0px|0|0px|false|false" animation_style="slide" animation_direction="top" animation_intensity_slide="2%" locked="off"][et_pb_row _builder_version="3.25" background_size="initial" background_position="top_left" background_repeat="repeat" custom_margin="|||" custom_padding="27px|0px|27px|0px" custom_width_px="1280px"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_text_color="#474ab6" text_line_height="1.9em" background_size="initial" background_position="top_left" background_repeat="repeat" text_orientation="center" max_width="540px" module_alignment="center" locked="off"]Let \( \{a_n\}_{n\ge 1} \) be a sequence of real numbers such that $$ a_n = \frac{1 + 2 + ... + (2n-1)}{n!} , n \ge 1 $$ . Then \( \sum_{n \ge 1 } a_n \) converges to ____________
[/et_pb_text][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" admin_label="Blog" _builder_version="3.22" custom_margin="|||" custom_padding="0px|0px|21px|0px|false|false"][et_pb_row _builder_version="3.25" background_size="initial" background_position="top_left" background_repeat="repeat" max_width="960px" custom_padding="0|0px|24px|0px|false|false" use_custom_width="on" custom_width_px="960px"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_tabs _builder_version="3.12.2"][et_pb_tab title="Hint 1 - Sum of odds" _builder_version="3.12.2"]Notice that \( 1 + 3 + 5 + ... + (2n-1) = n^2 \). A quick way to remember this is sum of first n odd numbers is \( n^2 \) Hence \( a_n = \frac{n^2}{n!} \) Simplifying this a little we have \( a_n = \frac {n}{(n-1)!} \)
[/et_pb_tab][et_pb_tab title="Hint 2 - Break in partials" _builder_version="3.12.2"]Notice that n = n-1 +1 Then \( a_n = \frac { n-1}{(n-1)!} + \frac {1}{ (n-1)!} = \frac { 1}{(n-2)!} + \frac {1}{ (n-1)!} \) We are interested in the limit of the partial sums of \(a_n \)
[/et_pb_tab][et_pb_tab title="Hint 3 - Something goes to e!" _builder_version="3.12.2"]Recall the \( \sum \frac {1}{n!} \) converges to the Euler number e. (Do not know this? Look up the definition of e in the internet). Here we have two copies of this series if we sum \( a_n \). (Since we are taking limit of a sum, (n-1) and (n-2) does not matter). Hence the limit of this sum is 2e,
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