Chapter 7: Q. 66 (page 641)

Use the principle of mathematical induction to prove that if $a_{k + 1} < a_{k} r$ for every k ≥ N, then $a_{N + n} < a_{N} r^{n}$ . Proving this implication completes our proof of the ratio test.

Short Answer

Expert verified

Hence, proved.

Step by step solution

Step 1. Given Information.

Given $a_{k + 1} < a_{k} r f o r e v e r y k ⩾ N .$

Step 2. Proof.

Let suppose it is true for k=N, then

$a_{N + 1} < a_{N} r .$

For k=N+1, we get:

$a_{N + 2} < a_{N + 1} r, a_{N + 2} < (a_{N} r) r a_{N + 2} < a_{N} r^{2} .$

The result is true for k=N.

Assume that the result is true for k=N+n-1,

$a_{N + n} < a_{N} r^{n} .$

Now we have to prove it for k=N+n.

Step 3. Proof part 2.

$F o r k = N + n, a_{N + n + 1} < a_{N + n} r, a_{N + n + 1} < (a_{N} r^{n}) r, f r o m s t e p 2 . a_{N + n + 1} < a_{N} r^{n + 1}, H e n c e, t h e r e s u l t i s t r u e f o r k = N + n . H e n c e b y t h e p r i n c i p l e o f m a t h e m a t i c a l i n d u c t i o n, r e s u l t i s t r u e f o r e v e r y p o s i t i v e n u m b e r s k ⩾ N .$

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Use the principle of mathematical induction to prove that if $a_{k + 1} < a_{k} r$ for every k ≥ N, then $a_{N + n} < a_{N} r^{n}$ . Proving this implication completes our proof of the ratio test.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Proof.

Step 3. Proof part 2.

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Use the principle of mathematical induction to prove that if ak+1<akrfor every k ≥ N, then aN+n<aNrn. Proving this implication completes our proof of the ratio test.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Proof.

Step 3. Proof part 2.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

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Study anywhere. Anytime. Across all devices.

Use the principle of mathematical induction to prove that if $a_{k + 1} < a_{k} r$ for every k ≥ N, then $a_{N + n} < a_{N} r^{n}$ . Proving this implication completes our proof of the ratio test.