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Consider the sequence {1k}k=0. The associated sequence {Sn}n=0, where

Sn=1+1+12!+...+1n!

, is a sequence of sums. In Chapter 8 we will see that this sequence converges to the number e. Evaluate Sn for n=1,2,3,10. How close is S10 to e?

Short Answer

Expert verified

For the sequence Sn=1+1+12!+...+1n!:

S1=2,

S2=2.5,

S3=2.667,

S10=2.71858.

The value of S10 is approximately 0.0003 more than the e.

Step by step solution

01

Step 1. Given Information

It is given that Sn=1+1+12!+...+1n!, is a sequence of sums.

The objective is to evaluate Sn for n=1,2,3,10 and to know how close is S10 to e.

02

Step 2. Find S1

For n=1 the sequence Sn=1+1+12!+...+1n! can be evaluated as:

S1=1+11!=1+1=2

03

Step 3. Find S2

For n=2 the sequence Sn=1+1+12!+...+1n! can be evaluated as:

S2=1+1+12!=1+1+0.5=2.5

04

Step 4. Find S3

For n=3 the sequence Sn=1+1+12!+...+1n! can be evaluated as:

S3=1+1+12!+13!=1+1+0.5+162.5+0.167=2.667

05

Step 5. Find S10

For n=10 the sequence Sn=1+1+12!+...+1n! can be evaluated as:

S10=1+1+12!+13!+14!+15!+16!+17!+18!+19!+110!=2+0.5+16+124+1120+1720+15040+140320+1362880+136288002.5+0.167+0.04167+0.0083+0.001389+0.000198+0.000025+0.00000276+0.000000276=2.71858

06

Step 6. How close is S10 to e.

The value of S10 is 2.71858 and the value of e is approximately 2.17828.

S10-e=2.71858-2.71828=0.0003

So, S10 is very close to e.

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