Question involving approximation, taylor series and provingQuestion: Consider the approximationln⁡(2)≈2(13+13×33+15×35)Prove that the error in this approximation is less than17×22×35Attempt: It looks like the expression comes from the taylor series expansion so: ln⁡(1+x)=x−x22+x33−x44+… for −1&lt;x&lt;1ln⁡(1−x)=−x−x22−x33−x44+…∴ln⁡(1+x1−x)=2(x+x33+x55+x77)Now let x=13∴ln⁡(2)=2(13+13×33+15×35+17×37)So we have to prove that:2(17×37+19×39+111×311⋯)&lt;17×22×35

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Question involving approximation, taylor series and provingQuestion: Consider the approximationln⁡(2)≈2(13+13×33+15×35)Prove that the error in this approximation is less than17×22×35Attempt: It looks like the expression comes from the taylor series expansion so: ln⁡(1+x)=x−x22+x33−x44+…  for   −1&amp;lt;x&amp;lt;1ln⁡(1−x)=−x−x22−x33−x44+…∴ln⁡(1+x1−x)=2(x+x33+x55+x77)Now let x=13∴ln⁡(2)=2(13+13×33+15×35+17×37)So we have to prove that:2(17×37+19×39+111×311⋯)&amp;lt;17×22×35

ostijum8dd · Accepted Answer

We can say that2(17×37+19×39+111×311⋯)&amp;lt;2(17×37+17×39+17×311⋯)=2(17×37)÷(1-132)=17×22×35

Janessa Foster · Accepted Answer

Since 9&amp;gt;7, and 11&amp;gt;7, and 13&amp;gt;7, and so on, the tail27⋅37+29⋅39+211⋅311+213⋅313+⋯is less than the sum of the geometric series27⋅37(1+132+134+136+⋯).But1+132+134+136+⋯=98.

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