Proving convergence of sequence with fraction sums [closed]

How do I prove convergence of this sequence (or prove that it doesn't exist)?

$$a_n=frac1{4+log 1}+frac1{4^2+log 2}+frac1{4^3+log 3}+ldots+frac1{4^n+log n}$$

edited Nov 29 '18 at 21:06

gimusi

asked Nov 29 '18 at 21:00

syd

closed as off-topic by T. Bongers, Scientifica, KReiser, Paul Plummer, Alexander Gruber♦ Nov 30 '18 at 3:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Scientifica, KReiser, Paul Plummer, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

The sequence is bounded and monotone increasing therefore it is convergent.
– MotylaNogaTomkaMazura
Nov 29 '18 at 21:06

add a comment |

How do I prove convergence of this sequence (or prove that it doesn't exist)?

$$a_n=frac1{4+log 1}+frac1{4^2+log 2}+frac1{4^3+log 3}+ldots+frac1{4^n+log n}$$

edited Nov 29 '18 at 21:06

gimusi

asked Nov 29 '18 at 21:00

syd

closed as off-topic by T. Bongers, Scientifica, KReiser, Paul Plummer, Alexander Gruber♦ Nov 30 '18 at 3:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Scientifica, KReiser, Paul Plummer, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

The sequence is bounded and monotone increasing therefore it is convergent.
– MotylaNogaTomkaMazura
Nov 29 '18 at 21:06

add a comment |

How do I prove convergence of this sequence (or prove that it doesn't exist)?

$$a_n=frac1{4+log 1}+frac1{4^2+log 2}+frac1{4^3+log 3}+ldots+frac1{4^n+log n}$$

edited Nov 29 '18 at 21:06

gimusi

asked Nov 29 '18 at 21:00

syd

How do I prove convergence of this sequence (or prove that it doesn't exist)?

$$a_n=frac1{4+log 1}+frac1{4^2+log 2}+frac1{4^3+log 3}+ldots+frac1{4^n+log n}$$

sequences-and-series

edited Nov 29 '18 at 21:06

gimusi

asked Nov 29 '18 at 21:00

syd

edited Nov 29 '18 at 21:06

gimusi

asked Nov 29 '18 at 21:00

syd

edited Nov 29 '18 at 21:06

gimusi

edited Nov 29 '18 at 21:06

gimusi

edited Nov 29 '18 at 21:06

gimusi

asked Nov 29 '18 at 21:00

syd

asked Nov 29 '18 at 21:00

syd

asked Nov 29 '18 at 21:00

syd

closed as off-topic by T. Bongers, Scientifica, KReiser, Paul Plummer, Alexander Gruber♦ Nov 30 '18 at 3:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Scientifica, KReiser, Paul Plummer, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by T. Bongers, Scientifica, KReiser, Paul Plummer, Alexander Gruber♦ Nov 30 '18 at 3:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Scientifica, KReiser, Paul Plummer, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

The sequence is bounded and monotone increasing therefore it is convergent.
– MotylaNogaTomkaMazura
Nov 29 '18 at 21:06

add a comment |

The sequence is bounded and monotone increasing therefore it is convergent.
– MotylaNogaTomkaMazura
Nov 29 '18 at 21:06

The sequence is bounded and monotone increasing therefore it is convergent.
– MotylaNogaTomkaMazura
Nov 29 '18 at 21:06

add a comment |

2 Answers
2

active

oldest

votes

HINT

We have that

$$left(frac1{4^n+log n}right)sim frac1{4^n}$$

answered Nov 29 '18 at 21:05

gimusi

add a comment |

-1

Your sequence is increasing :

$$a_{n+1}-a_n=frac{1}{4^{n+1}+ln(n+1)}>0.$$

on the other hand,

$$a_n=sum_{k=1}^nfrac{1}{4^k+ln(k)}le sum_{k=1}^nfrac{1}{4^k}$$

$$implies a_nle frac 14frac{1-frac{1}{4^n}}{1-frac 14}$$

$$implies a_nle frac 13$$
Hence, it is convergent.

edited Nov 30 '18 at 13:07

answered Nov 29 '18 at 21:11

hamam_Abdallah

38k21634

Thank you very much! So we're comparing it with a geometrical series which sum we know and then it must be smaller than that. It's that the approach we usually take when we deal with sequences given with sums of fractions?
– syd
Nov 29 '18 at 21:24

@syd Yes, we compare with something we know.
– hamam_Abdallah
Nov 29 '18 at 21:26

add a comment |

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

HINT

We have that

$$left(frac1{4^n+log n}right)sim frac1{4^n}$$

answered Nov 29 '18 at 21:05

gimusi

add a comment |

HINT

We have that

$$left(frac1{4^n+log n}right)sim frac1{4^n}$$

answered Nov 29 '18 at 21:05

gimusi

add a comment |

HINT

We have that

$$left(frac1{4^n+log n}right)sim frac1{4^n}$$

answered Nov 29 '18 at 21:05

gimusi

HINT

We have that

$$left(frac1{4^n+log n}right)sim frac1{4^n}$$

answered Nov 29 '18 at 21:05

gimusi

answered Nov 29 '18 at 21:05

gimusi

answered Nov 29 '18 at 21:05

gimusi

answered Nov 29 '18 at 21:05

gimusi

add a comment |

-1

Your sequence is increasing :

$$a_{n+1}-a_n=frac{1}{4^{n+1}+ln(n+1)}>0.$$

on the other hand,

$$a_n=sum_{k=1}^nfrac{1}{4^k+ln(k)}le sum_{k=1}^nfrac{1}{4^k}$$

$$implies a_nle frac 14frac{1-frac{1}{4^n}}{1-frac 14}$$

$$implies a_nle frac 13$$
Hence, it is convergent.

edited Nov 30 '18 at 13:07

answered Nov 29 '18 at 21:11

hamam_Abdallah

38k21634

Thank you very much! So we're comparing it with a geometrical series which sum we know and then it must be smaller than that. It's that the approach we usually take when we deal with sequences given with sums of fractions?
– syd
Nov 29 '18 at 21:24

@syd Yes, we compare with something we know.
– hamam_Abdallah
Nov 29 '18 at 21:26

add a comment |

-1

Your sequence is increasing :

$$a_{n+1}-a_n=frac{1}{4^{n+1}+ln(n+1)}>0.$$

on the other hand,

$$a_n=sum_{k=1}^nfrac{1}{4^k+ln(k)}le sum_{k=1}^nfrac{1}{4^k}$$

$$implies a_nle frac 14frac{1-frac{1}{4^n}}{1-frac 14}$$

$$implies a_nle frac 13$$
Hence, it is convergent.

edited Nov 30 '18 at 13:07

answered Nov 29 '18 at 21:11

hamam_Abdallah

38k21634

Thank you very much! So we're comparing it with a geometrical series which sum we know and then it must be smaller than that. It's that the approach we usually take when we deal with sequences given with sums of fractions?
– syd
Nov 29 '18 at 21:24

@syd Yes, we compare with something we know.
– hamam_Abdallah
Nov 29 '18 at 21:26

add a comment |

-1

Your sequence is increasing :

$$a_{n+1}-a_n=frac{1}{4^{n+1}+ln(n+1)}>0.$$

on the other hand,

$$a_n=sum_{k=1}^nfrac{1}{4^k+ln(k)}le sum_{k=1}^nfrac{1}{4^k}$$

$$implies a_nle frac 14frac{1-frac{1}{4^n}}{1-frac 14}$$

$$implies a_nle frac 13$$
Hence, it is convergent.

edited Nov 30 '18 at 13:07

answered Nov 29 '18 at 21:11

hamam_Abdallah

38k21634

Your sequence is increasing :

$$a_{n+1}-a_n=frac{1}{4^{n+1}+ln(n+1)}>0.$$

on the other hand,

$$a_n=sum_{k=1}^nfrac{1}{4^k+ln(k)}le sum_{k=1}^nfrac{1}{4^k}$$

$$implies a_nle frac 14frac{1-frac{1}{4^n}}{1-frac 14}$$

$$implies a_nle frac 13$$
Hence, it is convergent.

edited Nov 30 '18 at 13:07

answered Nov 29 '18 at 21:11

hamam_Abdallah

38k21634

edited Nov 30 '18 at 13:07

answered Nov 29 '18 at 21:11

hamam_Abdallah

38k21634

answered Nov 29 '18 at 21:11

hamam_Abdallah

38k21634

answered Nov 29 '18 at 21:11

hamam_Abdallah

38k21634

Thank you very much! So we're comparing it with a geometrical series which sum we know and then it must be smaller than that. It's that the approach we usually take when we deal with sequences given with sums of fractions?
– syd
Nov 29 '18 at 21:24

@syd Yes, we compare with something we know.
– hamam_Abdallah
Nov 29 '18 at 21:26

add a comment |

Thank you very much! So we're comparing it with a geometrical series which sum we know and then it must be smaller than that. It's that the approach we usually take when we deal with sequences given with sums of fractions?
– syd
Nov 29 '18 at 21:24

@syd Yes, we compare with something we know.
– hamam_Abdallah
Nov 29 '18 at 21:26

Thank you very much! So we're comparing it with a geometrical series which sum we know and then it must be smaller than that. It's that the approach we usually take when we deal with sequences given with sums of fractions?
– syd
Nov 29 '18 at 21:24

@syd Yes, we compare with something we know.
– hamam_Abdallah
Nov 29 '18 at 21:26

add a comment |

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