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 #2
avatar+61 
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Feb 6, 2017
 #1
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Feb 6, 2017
 #8
avatar+26388 
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Hi ! I have to show that
\(\sum \limits_{n=2}^{\infty} \left( \dfrac { \ln \left[(1+ \frac {1} {n})^n\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \right)= \dfrac {1} {2\cdot \ln(2)}\)
Can anyone help ? Thank you in advance.

 

\(\begin{array}{|rcll|} \hline && \sum \limits_{n=2}^{\infty} \dfrac { \ln \left[(1+ \frac {1} {n})^n\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln \left[(\frac {n+1} {n})^n\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln \left[ \frac {(n+1)^n} {n^n}\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln [ \frac {(n+1)^{n+1}} {n^n} ]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln \left[(n+1)^{n+1} \right]- \ln(n^n) } { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right] } \\\\ &=&\sum \limits_{n=2}^{\infty} \left( \dfrac { \ln \left[(n+1)^{n+1} \right] } { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right] } - \dfrac { \ln(n^n) } { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right] } \right) \\\\ &=&\sum \limits_{n=2}^{\infty} \left( \dfrac { 1 } { \ln(n^n) } - \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \right) \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln(n^n) } -\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \\\\ &=&\sum \limits_{n=1}^{\infty} \dfrac { 1 } { \ln[(n+1)^{n+1}] } -\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \\\\ &=& \dfrac{1}{\ln(2^2)} + \sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln[(n+1)^{n+1}] } -\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \\\\ &=& \dfrac{1}{\ln(2^2)} \\\\ &=& \dfrac{1}{2\cdot \ln(2)} \\ \hline \end{array}\)

 

laugh

Feb 6, 2017
 #2
avatar+9665 
0

We have less things to do....... Hong Kong have a different version of geometric proof......

 

Let me do this in my version.

 

\(\begin{array}{lr}1.\angle\text{BAD}=\angle\text{CAD}&\text{angle bisector}\\2.\angle\text{BAD}=\angle\text{CFD}&\text{alt.}\angle s,AB//CF\\\quad\therefore\angle \text{CAD} = \angle \text{CFD}\\ 3.\text{CA}=\text{FC}&\text{sides. opp.,eq. }\angle s\\ 4.\angle\text{ADB}=\angle\text{CDF}&\text{vert. opp. }\angle s\\ 5.\angle\text{ABD}=\angle\text{FCD}&\text{alt.}\angle s,\;AB//CF\\ 6.\triangle\text{ABD}\sim\triangle\text{FCD}&\text{A.A.A.}\\ 7.\dfrac{\text{AB}}{\text{BD}}=\dfrac{\text{FC}}{\text{CD}}&\text{corr. sides,}\sim \triangle s\\ 8.\dfrac{\text{AB}}{\text{BD}}=\dfrac{\text{AC}}{\text{CD}}\end{array}\)

 

We don't need to write the below stuff but it is just a explanation for you.

 

\(\text{alt. }\angle s,\text{AB}//\text{CD}\) means the alternate interior angles of the 2 parallel lines AB and CD. (Yes we use // instead of ||)We don't have alternate exterior angles so if you write alt. angles it must mean alternate interior angles. (So that the proof for alternate exterior angles are more complicated. LOL)

 

The reason "angle bisector" just means "definition of angle bisectors".

 

sides opp., eq angles means when the base angles are equal, that triangle is a isosceles triangle. Don't ask me why "opp.". I don't know. My teacher neither.

 

A.A.A. is the Angle Angle Similarity Postulate. A.A.A. means Angle Angle Angle. We have to prove both 3 angles are equal to prove similarity.

 

corr. sides similar triangles means "the corresponding sides of similar triangles are equal."

 

As you see:

 

1) We use \(=\) instead of \(\cong\) for angles because the congruent symbol for representing triangle congruency. And we use // instead of ||

 

2) We have different reasons.

 

3) We do not need to write any reasons for some lines. No reason is just blank. You don't need to put something like 'Substitution Property'.

Feb 6, 2017

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