Sn=(1)/(1+2 i)-(1)/(2+2 i)+(1)/(2+2 i)-(1)/(3+2 i)+(1)/(3+2 i)-(1)/(4+2 i)+⋯+(1)/(n+2 i)-(1)/

Name: Problem 13, Chapter 19: Advanced Engineering Mathematics
Uploaded: 2021-08-27T14:09:33-08:00
Duration: 6 min 12 s
Channel: Sandip Ranjan
Description: Sn=(1)/(1+2 i)-(1)/(2+2 i)+(1)/(2+2 i)-(1)/(3+2 i)+(1)/(3+2 i)-(1)/(4+2 i)+⋯+(1)/(n+2 i)-(1)/(n+1+2 i)=(1)/(1+2 i)-(1)/(n+1+2 i) Thus, limn →∞ Sn=(1)/(1+2 i)=(1)/(5)-(2)/(5) i

step 1 In this question, we have to find the value of limit n tends to infinity into 1 .1 minus n squar

Question

$$\begin{aligned}&S_{n}=\frac{1}{1+2 i}-\frac{1}{2+2 i}+\frac{1}{2+2 i}-\frac{1}{3+2 i}+\frac{1}{3+2 i}-\frac{1}{4+2 i}+\cdots+\frac{1}{n+2 i}-\frac{1}{n+1+2 i}=\frac{1}{1+2 i}-\frac{1}{n+1+2 i}\\&\text { Thus, } \lim _{n \rightarrow \infty} S_{n}=\frac{1}{1+2 i}=\frac{1}{5}-\frac{2}{5} i \end{aligned}$$

   $$\begin{aligned}&S_{n}=\frac{1}{1+2 i}-\frac{1}{2+2 i}+\frac{1}{2+2 i}-\frac{1}{3+2 i}+\frac{1}{3+2 i}-\frac{1}{4+2 i}+\cdots+\frac{1}{n+2 i}-\frac{1}{n+1+2 i}=\frac{1}{1+2 i}-\frac{1}{n+1+2 i}\\&\text { Thus, } \lim _{n \rightarrow \infty} S_{n}=\frac{1}{1+2 i}=\frac{1}{5}-\frac{2}{5} i
\end{aligned}$$

Advanced Engineering Mathematics

Dennis G. Zill,… 3rd Edition

Chapter 19, Problem 13

Instant Answer

Solved by Expert Sandip Ranjan

08/27/2021

Step 1

Specifically, the negative term in each pair cancels out with the positive term in the next pair. Show more…

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$$\begin{aligned}&S_{n}=\frac{1}{1+2 i}-\frac{1}{2+2 i}+\frac{1}{2+2 i}-\frac{1}{3+2 i}+\frac{1}{3+2 i}-\frac{1}{4+2 i}+\cdots+\frac{1}{n+2 i}-\frac{1}{n+1+2 i}=\frac{1}{1+2 i}-\frac{1}{n+1+2 i}\\&\text { Thus, } \lim _{n \rightarrow \infty} S_{n}=\frac{1}{1+2 i}=\frac{1}{5}-\frac{2}{5} i \end{aligned}$$

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Key Concepts

Series Convergence

Series convergence involves determining whether the sum of an infinite sequence of terms approaches a finite limit. Analyzing convergence, particularly through the limit of the partial sums, is critical in ensuring that an infinite telescoping series yields a valid finite result.

Complex Numbers

Complex numbers, expressed in the form a + bi where i is the imaginary unit, require special arithmetic treatment. Manipulating these numbers often involves performing operations such as addition, subtraction, and finding reciprocals, which is essential in handling series that involve complex denominators.

Telescoping Series

A telescoping series is one in which most terms cancel out sequentially, leaving only a few terms from the beginning and end of the series. This characteristic cancellation simplifies the evaluation of the series' sum by focusing solely on the remaining boundary terms after cancellation.

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