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The question:

a) Differentiate both sides of the geometric series with respect to r:

$\sum_{i=0}^n r^i=\frac{1-r^{n+1}}{1-r}$

b) Use the result in part (a) to show that

$\sum_{i=1}^nir^i < \frac{r}{(1-r)^2}$

for all n≥1.

Solution for a) Quotient rule.

$\sum_{i=0}^nir^{i-1}=\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2}$

b) With mathematical induction:

$P(n): \sum_{i=1}^nir^i < \frac{r}{(1-r)^2}$

$n=1, LHS = r^2 - 2r + 1, RHS = 2r^2$

Assume P(k) is true

$P(k): \sum_{i=1}^kir^i < \frac{r}{(1-r)^2}$

Prove P(k+1) is true:

$P(k+1): \sum_{i=1}^{k+1}ir^i < \frac{r}{(1-r)^2}$

$\sum_{i=1}^kir^i+(k+1)*ir^(k+1) < \frac{r}{(1-r)^2}$

Stuck here, please enlighten

Thank you, Best regards

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    Images of text are discouraged on this site as they don't allow other users to find this question (because search engines can't read images of text). Also, questions with no effort put into them aren't appropriate for this site. Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 2017-02-02
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    What is P(K+1)? – 2017-02-02
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    Thank you for your comments, first user and I have tried to edit and included my step, May I ask for your helps please =) thanks. – 2017-02-02
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    why the downvotes for the poor guy who included his steps? – 2017-02-02

1 Answers 1

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Assume $0

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    Dear Mr Michel Gonnord, Thank you for your response, I think I am expected to proof it with mathematical induction, May I ask if you would have any hints about it? – 2017-02-02