Evaluating an infinite series ∑k=1∞(43k−43k+1) two ways. a. Use a telescoping series argument. b. Use a geometric series argument.

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Evaluating an infinite series ∑k=1∞(43k−43k+1) two ways.  a. Use a telescoping series argument.  b. Use a geometric series argument.

Nicole Conner · Accepted Answer

Consider the given series:  ∑un=∑k=1∞(43k−43k+1)  Here the objective is to evaluate the given series using  Telescoping series argument  Geometric series argument.  Telescoping series argument  Telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. The cancellation technique, with part of each term canceling with part of the next term, is known as the method of differences.  Here  ∑un=∑k=1∞(43k−43k+1)  ∑un=∑k=1∞43k−∑k=1∞43k+1  Expand the series by substituting k=0,1,2,3…  ∑un=(431+432+433+434+435+436+…)−(432+433+434+435+436+437+…)  ∑un=43  Geometric series argument  An infinite series of the form  S=a+ar+ar2+ar3+…∞  S=a1−r  Consider the given series  ∑un=∑k=1∞(43k−43k+1)  ∑un=∑k=1∞43k(1−13)  ∑un=∑k=1∞43k(23)  ∑un=(83)∑k=1∞13k

Evaluating an infinite series sum_{k=1}^infty(frac{4}{3^k}-frac{4}{3^{k+1}}) two ways. a. Use a telescoping series argument. b. Use a geometric series argument.

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