Q.1 If鈭慿=1鈭瀉k聽and 鈭慿=1鈭瀊k聽... [FREE SOLUTION]

Chapter 7: Q.1 (page 625)

If ${鈭�}_{k = 1}^{鈭�} a_{k}$ and ${鈭�}_{k = 1}^{鈭�} b_{k}$ are convergent series and c is any real number ,then
a) ${鈭�}_{k = 1}^{鈭�} c a_{k} = c {鈭�}_{k = 1}^{鈭�} a_{k}$
b) ${鈭�}_{k = 1}^{鈭�} (a_{k} + b_{k}) = {鈭�}_{k = 1}^{鈭�} a_{k} + {鈭�}_{k = 1}^{鈭�} b_{k}$

Short Answer

Expert verified

This means that adding or removing any finite number of terms from the start of the series does not change its convergent series

Step by step solution

the given data

A) since we already know that both limits and sums commute with sums and constant multiples we will prove part b and leave the proof of part a )

B) given information ∑k=1∞(ak+bk)=∑k=1∞ak+∑k=1∞bk

The result follows from looking at the limits of partial sums

${鈭�}_{k = 1}^{鈭�} (a_{k} + b_{K}) = \lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{鈭�} (a_{k} + b_{k}) = \lim_{n 鈫� 鈭�} ({鈭�}_{k = 1}^{鈭�} a_{k} + {鈭�}_{k = 1}^{鈭�} b_{k}) = \lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{鈭�} a_{k} + \lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{鈭�} b_{k} = {鈭�}_{k = 1}^{鈭�} a_{k} + {鈭�}_{k = 1}^{鈭�} b_{k}$

hence proved

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Short Answer

Step by step solution

the given data

B) given information ∑k=1∞(ak+bk)=∑k=1∞ak+∑k=1∞bk

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91影视

If鈭�k=1鈭�akand 鈭�k=1鈭�bkare convergent series and c is any real number ,thena)鈭�k=1鈭�cak=c鈭�k=1鈭�akb)鈭�k=1鈭�(ak+bk)=鈭�k=1鈭�ak+鈭�k=1鈭�bk

Short Answer

Step by step solution

the given data

B) given information ∑k=1∞(ak+bk)=∑k=1∞ak+∑k=1∞bk

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Logic and Functions

Probability and Statistics

Pure Maths

Calculus

Statistics

Study anywhere. Anytime. Across all devices.